Thermodynamic properties of a 2-dim ideal gas

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

\displaystyle  S=Nk\left[\ln\frac{2\pi mAU}{\left(hN\right)^{2}}+2\right] \ \ \ \ \ (1)

where {A} is the area occupied by the gas, {N} is the number of molecules, each of mass {m}, and {U} 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 {V} by the area {A}):

\displaystyle  dU=TdS-PdA+\mu dN \ \ \ \ \ (2)

The temperature is determined from the entropy as

\displaystyle   \frac{1}{T} \displaystyle  = \displaystyle  \left(\frac{\partial S}{\partial U}\right)_{A,N}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  Nk\frac{\left(hN\right)^{2}}{2\pi mAU}\frac{2\pi mA}{\left(hN\right)^{2}}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{Nk}{U} \ \ \ \ \ (5)

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

\displaystyle  U=\frac{2}{2}NkT=NkT \ \ \ \ \ (6)

The pressure can be obtained from

\displaystyle   P \displaystyle  = \displaystyle  T\left(\frac{\partial S}{\partial A}\right)_{U,N}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  Nk\frac{\left(hN\right)^{2}}{2\pi mAU}\frac{2\pi mU}{\left(hN\right)^{2}}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \frac{NkT}{A} \ \ \ \ \ (9)

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

\displaystyle  PA=NkT \ \ \ \ \ (10)

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

\displaystyle   \mu \displaystyle  \equiv \displaystyle  -T\left(\frac{\partial S}{\partial N}\right)_{U,A}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  -kT\left[\ln\frac{2\pi mAU}{\left(hN\right)^{2}}+2\right]-NkT\left(-\frac{2}{N}\right)\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  -kT\ln\frac{2\pi mAU}{\left(hN\right)^{2}}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  -kT\ln\left(\frac{A}{N}\frac{2\pi mkT}{h^{2}}\right) \ \ \ \ \ (14)

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

\displaystyle  \mu=-kT\ln\left[\frac{V}{N}\left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}\right] \ \ \ \ \ (15)

The only differences are the replacement of {V} by {A} and the change in the exponent inside the logarithm from {\frac{3}{2}} 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 {3N} degrees of freedom, while for a 2-d gas, there are {2N}. Thus the exponent in the 2-d case is {\frac{2}{3}} 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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