Introduction to Mathematical Physics/N body problems and statistical equilibrium/Thermodynamical perfect gas

In this section, a perfect gas model is presented: all the particles are independent, without any interaction.

Remark:

A perfect gas corresponds to the case where the kinetic energy of the particles is large with respect to the typical interaction energy. Nevertheless, collisions between particles that can be neglected are necessary for the thermodynamical equilibrium to establish.

Classical approximation (see section secdistclassi) allows to replace the sum over the quantum states by an integral of the exponential of the classical hamiltonian $H(q_{i},p_{i})$. The price to pay is just to take into account a proportionality factor ${\frac {1}{2\pi \hbar }}$. Partition function $z$ associated to one particle is:

$z={\frac {1}{2\pi \hbar }}\int dq_{i}dp_{i}e^{-\beta H(p,q)}$

$z={\frac {1}{2\pi \hbar }}\int dq^{3}\int dp^{3}e^{-\beta H(p,q)}$

Partition function $z$ is thus proportional to $V$ :

$z=A(\beta )V$

Because particles are independent, partition function $Z$ for the whole system can be written as:

$Z={\frac {1}{N!}}z^{N}$

It is known that pressure (proportional to the Lagrange multiplier associated to the internal variable "volume") is related to the natural logarithm of $Z$; more precisely if one sets:

$F=-k_{B}T\log Z$

then

$p=-{\frac {\partial F}{\partial V}}.$

This last equation and the expression of $Z$ leads to the famous perfect gas state equation:

$pV=Nk_{B}T$