1 Kinetics 1.1 Big numbers and probability 1.3 The Boltzmann factor

1.2 Estimates of ideal gasses and the equipartition theorem

(September 1, 2023)

The pressure of an ideal gas satisfies

pV=n_{\rm ml}RT

(1.25)

Here $n_{\rm ml}$ is the number of moles, which is the number of particles $N$ in units of Avogadro’s number, $n_{\rm ml}\equiv N/N_{A}$ . The symbol $n\equiv N/V$ is reserved for the number of particles per volume:

\mbox{ $n\equiv\frac{N}{V}$ \qquad{\bf is NOT} \qquad$n_{\rm ml}\equiv\frac{N}% {N_{A}}$ }\,.

(1.26)

We will work with the number of particles $N$ instead of $n_{\rm ml}$ and define Boltzmann’s constant $k_{B}$

pV=N{k_{\scriptscriptstyle B}T}\qquad k_{B}\equiv\frac{R}{N_{A}}

(1.27)

Sometimes we will drop the “B” and just write ${kT}$ for ${k_{\scriptscriptstyle B}T}$ .

Numerically

R=8.32\,{\rm J}/{}^{\circ}{\rm K}\qquad k_{B}=\frac{\tfrac{1}{40}eV}{300{}^{% \circ}{\rm K}}

(1.28)

The reason for writing $k_{B}$ like this is because this is how people (including me) remember it: e.g. typical thermal energy, $\sim k_{B}T$ , is “one fortieth of electron volt at room temperature”, $T\simeq 300{}^{\circ}{\rm K}$ .

The typical value of pressure is $1\,{\rm bar}=10^{5}{\rm N}/{\rm m}^{2}\simeq 1\,{\rm atm}$ , a typical volume is a liter, $1\,{\rm L}=(10{\rm cm})^{3}=1000\,{\rm cm}^{3}=10^{-3}{\rm m}^{3}$ . We note

(1\,{\rm bar})\,(1\,{\rm L})=100\,{\rm J}

(1.29)

Standard Temperature and Pressure (STP) is one bar at $273{}^{\circ}{\rm K}$ (freezing). The volume of one mole of gas at STP is 22 L. Keep in mind that under STP N(particles), V(volume), and U(total energy) are Extensive which mean they will grow with the system size. While T(temperature) and P(pressure) are Intensive and are constant throughout.

The equi-partition theorem states that mean energy per “degree of freedom (dof)” in the gas is ${\textstyle\frac{1}{2}}{kT}$ . We will explain what we mean here by dof using examples. Take a mono-atomic gas. Each an atom which can move in three ways – in the $x$ , the $y$ , and the $z$ directions. Thus the number of dof is $3N$ where $N$ is the number of atoms in the gas. So the total mean total energy in the energy in the gas, which we call $U$ or $E$ (they are the same in our notation), is

U\equiv E=\frac{3}{2}N{kT}

(1.30)

The energy (or Hamiltonian) of each particle, which we typically call $\epsilon$ , is a sum of three quadratic forms

\epsilon=\frac{1}{2}mv_{x}^{2}+\frac{1}{2}mv_{y}^{2}+\frac{1}{2}mv_{z}^{2}=% \frac{1}{2}m\vec{v}^{2}

(1.31)

Technically, the equipartition theorem says that the mean energy of each independent subsystem (i.e. a single particle) is ${\textstyle\frac{1}{2}}k_{B}T$ per quadratic form in the classical Hamiltonian – there are three forms counting the $v_{x}^{2}$ , $v_{y}^{2}$ and $v_{z}^{2}$ terms. Each quadratic form gives ${\textstyle\frac{1}{2}}{k_{\scriptscriptstyle B}T}$ so

\left\langle\frac{1}{2}mv_{x}^{2}\right\rangle={\textstyle\frac{1}{2}}{k_{% \scriptscriptstyle B}T}

(1.32)

and

\left\langle\frac{1}{2}m\vec{v}^{2}\right\rangle=\frac{3}{2}{k_{% \scriptscriptstyle B}T}

(1.33)

The root means square velocity is

v_{\rm rms}=\sqrt{\left\langle\vec{v}^{2}\right\rangle}=\sqrt{\frac{3{k_{% \scriptscriptstyle B}T}}{m}}

(1.34)

and is typically a couple of hundred meters a second, i.e. close to the speed of sound $c_{s}\simeq 330\,{\rm m/s}$ .

For a classical diatomic gas there are five degrees of freedom (quadratic forms) per molecule, since the diatomic molecule can also rotate around the $x$ and $y$ axis. One must include the translational and rotational kinetic energy

\displaystyle\epsilon=

\displaystyle\frac{1}{2}mv_{x}^{2}+\frac{1}{2}mv_{y}^{2}+\frac{1}{2}mv_{z}^{2}% +\frac{1}{2}I\omega_{x}^{2}+\frac{1}{2}I\omega^{2}_{y}

(1.35)

We note that instead of working with the velocity and angular velocity we will increasingly work with the momentum $p_{x}=mv_{x}$ and the angular momentum $L_{x}=I\omega_{x}$ ,

	$\displaystyle\epsilon=$	$\displaystyle\frac{p_{x}^{2}}{2m}+\frac{p_{y}^{2}}{2m}+\frac{p_{z}^{2}}{2m}+% \frac{L_{x}^{2}}{2I}+\frac{L_{y}^{2}}{2I}$		(1.36)
	$\displaystyle=$	$\displaystyle\frac{\vec{p}^{2}}{2m}+\frac{\vec{L}^{2}}{2I}$		(1.37)

The average of each of these five quadratic forms is ${\textstyle\frac{1}{2}}k_{B}T$ so the mean energy per particle is

\frac{U}{N}=\frac{5}{2}{k_{\scriptscriptstyle B}T}

(1.38)

The formulas in this section can be used to make a variety of estimates such as: the spacing between particles at room temperature; the typical speed; the typical de Broglie wavelength; typical angular velocity and angular momentum. We defined the thermal de Broglie wavelength:

\lambda_{\rm th}=\frac{h}{\sqrt{2\pi m{k_{\scriptscriptstyle B}T}}}

(1.39)

We note that $\lambda_{\rm th}\sim h/mv_{\rm rms}$ . The factor of $\sqrt{2\pi}$ here is purely a matter of convention.