Problem: Consider placing a fictitious open surface in an equilibrium ideal gas at temperature \(T\); although the net particle current density through such a surface would be \(\textbf J=\textbf 0\), if one only counts the particles that go through the surface from one side to the other, then show that the resulting unidirectional particle current density \(J\) is non-zero, and given by the Hertz-Knudsen equation:
\[J=\frac{1}{4}n\langle v\rangle\]
where \(n=p/k_BT\) is the number density and \(\langle v\rangle=\sqrt{8k_BT/\pi m}\) the average speed.
Solution:
Problem: By an analogous calculation, show that the unidirectional kinetic energy current density \(S\) for an ideal gas (which one might also think of as a heat flux \(S=q\)) is given by:
\[S=\frac{1}{2}nk_BT\langle v\rangle\]
And hence, show that the average kinetic energy of particles hitting a wall is enhanced by a Bayes’ factor of \(4/3\) compared to the bulk kinetic energy \(\frac{3}{2}k_BT\) per particle.
Solution:
Problem: Using kinetic theory, obtain a simple expression for the Langmuir adsorption isotherm for the adsorbed fraction \(\theta(p)\) as a function of the pressure \(p\). Then, obtain the same result using statistical mechanics.
Solution: The Hertz-Knudsen equation can be rewritten as \(J=p/\sqrt{2\pi mk_BT}\) which shows that, for isothermal fixed \(T\), \(J\propto p\). On the basis of this, the adsorption rate per unit area is postulated to be of the form \(k_ap(1-\theta)\) whereas the desorption rate is \(p\)-independent and simply given by \(k_d\theta\). Equating the two yields the Langmuir adsorption isotherm (which depends only on the equilibrium constant \(K:=k_a/k_d\)):
\[\theta(p)=\frac{Kp}{1+Kp}\]
The statistical mechanical version of this argument is to consider a vapor in equilibrium with the surface, and in particular to equate the chemical potential \(k_BT\ln n\lambda_T^3\) of the vapor with that of the surface…(flesh this argument out later).