Classical ideal gas

The equation of states and energy

Let us now apply the above equations to rederive the ideal gas properties. Consider a box filled with $\mathcal{N}$ particles. These particles are free and non interacting. Treating them classicaly in the context of Newtonian mechanics, we can write the energy of each particle as $mv^2/2 = \vec{p}^2/2m$ (See the classical mechanics lecture on this topic). As such, the energy of a microstate, i.e. a particle configuration in the box is given by

\[E(i)=\frac{1}{2m}\sum_n^{\mathcal{N}}{\vec{p}_n}^2\]

where we sum over all $\mathcal{N}$ independent particles.

The partition function associated to the maximisation of entropy is hence

\[Z=\int e^{-\frac{\beta}{2m}\sum_n \vec{p}_n^2} d^{3\mathcal{N}}x d^{3\mathcal{N}}p\]

Where we replaced the sums over the $N$ possible microstates associated to the macrostate by continuous integrals over momentum and space, as discussed in the previous lecture.

Develloping this integral simply gives

\[\boxed{Z=\left(\frac{e}{\rho}\right)^\mathcal{N}\left(\frac{2m\pi}{\beta}\right)^{\frac{3\mathcal{N}}{2}}}\]

Proof

The integral over space gives $$\int dx^{3\mathcal{N}}=\frac{V^\mathcal{N}}{\mathcal{N}!}$$ where the $\mathcal{N}!$ is here to traduce the fact that particles are distinguishable and avoid double conting same particle configurations with different labeling. $$Z=\frac{V}{\mathcal{N}!}\left(\frac{2m\pi}{\beta}\right)^{\frac{3\mathcal{N}}{2}}$$

From $\langle E\rangle$, one can find back the mean energy of the macrostate under consideration to be

\[\boxed{U = \langle E \rangle = -\frac{\partial \ln(Z)}{\partial \beta} = \frac{3}{2}\mathcal{N}k_B T}\]

Proof

$$\begin{aligned} \ln(Z)&=\ln\left(\left(\frac{e}{\rho}\right)^N\left(\frac{2m\pi}{\beta}\right)^{\frac{3N}{2}}\right) \\ &= N\ln\left(\frac{e}{\rho}\right)+\frac{3N}{2}\left(\ln(2m\pi)-\ln(\beta)\right)\\ \end{aligned} $$ $$\begin{aligned} -\frac{\partial \ln(Z)}{\partial \beta}&= -\frac{3\mathcal{N}}{2}\frac{\partial}{\partial \beta}\left(-\ln(\beta)\right) \\ &= \frac{3\mathcal{N}}{2} \frac{1}{\beta}\\ &= \frac{3\mathcal{N}}{2} k_B T \end{aligned} $$

Similarly, we can compute the pressure of the macrostate to be

\[P = \frac{\partial{E}}{\partial{V}}\Bigg|_S= \rho k_B T\]

that is, we derived back the ideal gas law

\[\boxed{P =\frac{n\mathcal{R}T}{V}}\]

and thus, only by assuming that particles were not interacting i.e. that the energy of a microstate is given by the sum of the kinetic energy of each particle $E(i)=\sum^{\mathcal{N}}_k E_c(k)$, and asking for the entropy to be maximal!

Proof

As $$P= -\frac{\partial E}{\partial V}\Bigg|_{S}$$ and $$ E= \frac{3N}{2} k_B T= \frac{3\mathcal{N}}{2}k_B $$

The Maxwell-Boltzmann speed distribution

We can use the partition function to associate a probability to a given microstate

\[p_i = \frac{1}{Z}e^{-\beta E_i}\] \[p(E)\text{d}E = \frac{1}{Z}e^{-\beta E_i}\]

Single particle