Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problem 3.39.

We now revisit the 2-d ideal gas for which the Sackur-Tetrode equation is

where is the area occupied by the gas, is the number of molecules, each of mass , and is the total energy. We can work out the temperature, pressure and chemical potential by applying the thermodynamic identity adapted for 2 dimensions (by replacing the volume by the area ):

The temperature is determined from the entropy as

This just gives us the formula from the equipartition theorem for a system with 2 degrees of freedom:

The pressure can be obtained from

This is just the 2-dim analogue of the ideal gas law:

Finally, chemical potential is defined in terms of the entropy as

We can compare this to the chemical potential for a 3-d ideal gas

The only differences are the replacement of by and the change in the exponent inside the logarithm from to 1. The latter arises from the derivation of the multiplicity, where the exponent depends on the number of degrees of freedom in the system. For a 3-d gas, there are degrees of freedom, while for a 2-d gas, there are . Thus the exponent in the 2-d case is that in the 3-d case. [You’d need to follow through the derivation in detail to see the difference, but basically that’s where it comes from.]

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