The ideal gas law (the thermic equation of state for an ideal gas) is

$pV=N{k}_{B}T,$

(A.1)

where $N$ is the number of molecules of the gas, $k_B$ is the Boltzmann constant and $p$, $V$, $T$ are the state variables pressure, volume and temperature respectivly. We can rewrite this equation also in the following ways

	$pV$	$={\displaystyle \frac{N}{N_A}}{N}_A{k}_{B}T$
		$=nRT$
		$={\displaystyle \frac{m}{M}}RT$
	$\Rightarrow p$	$={\displaystyle \frac{m}{V}}{\displaystyle \frac{R}{M}}T$
		$=\rho R_{S}T$

where we used the number of moles $n=\frac{N}{N_A}$, the molar mass of a gas molecule $M=\frac{m}{n}$ ($m$ is the mass of the gas molecule), the universal gas constant $R=N_A{k}_B$ and the specific gas constant $R_S=\frac{R}{M}$.

From statistical thermodynamics it follows that the internal energy $U$ of an ideal gas is related to the temperature via the so called caloric equation of state

$U={\displaystyle \frac{f}{2}}N{k}_{B}T.$

(A.2)

$f$ specifies the number of degrees of freedom of a molecule (3 for a one-atomic gas, 5 for a two-atomic gas and 6 for a nonlinear many-atomic gas)²³²³These values are for a three dimensional world! If we would be in a two-dimensional world we would have 2 for a one-atomic gas, 3 for a two-atomic gas and 3 for any asymmetric many-atomic gas.. The adiabatic coefficient $\gamma$ is related to $f$ via

$\gamma ={\displaystyle \frac{c_p}{c_v}}=1+{\displaystyle \frac{2}{f}}$

(A.3)

where $\frac{c_p}{c_v}$ is the ratio of the specific heat capacity at constant pressure $c_p$ to the specific heat capacity at constant volume $c_v$ of an ideal gas.²⁴²⁴Is this relation dependent on the number of dimensions we live in?

Inserting (A.1) and (A.3) into (A.2) we get for the pressure of an ideal gas

$p=(\gamma -1)\rho u$

(A.4)

where $u$ is the specific internal energy. The specific internal energy can be computed from the total energy ²⁵²⁵This depends on the definition of the total energy! If one includes the potential energy in the total energy one has to substract also the potential energy to get the internal energy. via $u=e-\frac{1}{2}{v}^2$ and so we get for the pressure

$p=\left(\gamma -1\right)\rho \left(e-{\displaystyle \frac{1}{2}}{v}^2\right)$

(A.5)

…to be completed

…to be completed