Density of states of ideal gas and photon gas

In summary, the conversation is about deriving the density of states for an ideal gas and a photon gas. The ideal gas is described by the energy equation E_n = A*n^2, where A = (h_bar^2*pi^2)/(2mL^2). The question is whether the same energy equation can be used for a photon gas, described by E_n = (h_bar*pi*c/L)*n, or if the first equation should be used. The individual is confused about which equation to use and is seeking clarification.
  • #1
ausdreamer
23
0

Homework Statement



I know how to derive the density of states for an ideal gas by using the energy equation:

E_n = A*n^2, where A = (h_bar^2*pi^2)/(2mL^2)

but what about for a 'photon gas'? Do I use the same energy equation as above, or the following:

E_n = (h_bar*pi*c/L)*n

Homework Equations





The Attempt at a Solution



I think I'd use the second equation, but solutions to the assignment said use the first equation. I'm confused!
 
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  • #2
Regular gas in a box derivation.
Note: photons are Bosons and massless with two spin states.
 
Last edited:
  • #3
Thanks, this helped a lot :)
 

Related to Density of states of ideal gas and photon gas

1. What is the difference between the density of states of an ideal gas and a photon gas?

The density of states of an ideal gas refers to the number of available energy states for each particle in the gas. In contrast, the density of states of a photon gas refers to the number of available energy states for each photon in the radiation field.

2. How is the density of states of an ideal gas related to temperature?

The density of states of an ideal gas is directly proportional to temperature. As temperature increases, the particles in the gas have more energy and therefore more available energy states, leading to a higher density of states.

3. Does the density of states of a photon gas change with temperature?

Yes, the density of states of a photon gas is also directly proportional to temperature. As temperature increases, the photons have more energy and therefore more available energy states, leading to a higher density of states.

4. Can the density of states of an ideal gas or photon gas be zero?

No, the density of states of both an ideal gas and a photon gas cannot be zero. This is because there will always be some energy states available, even at very low temperatures.

5. How does the density of states of an ideal gas or photon gas affect the thermodynamic properties of the system?

The density of states plays a crucial role in determining the thermodynamic properties of a system. It affects the partition function, which is used to calculate quantities such as the internal energy, entropy, and free energy of the system. A higher density of states leads to a higher partition function and therefore higher thermodynamic properties.

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