# 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.]