Consider a rectangular block of wood moving with a velocity $v_{0}$ in a gas at temperature $T$ and mass density $\rho$. Assume the velocity is along the x-axis and the area of cross-section of the block perpendicular to $v_{0}$ is $A$. Show that the drag force on the block is $4\rho A v_{0} \sqrt{\frac{kT}{m}}$,where $m$ is the mass of the gas molecule.

(N/A) Let $n$ be the number density of gas molecules.
Let $v$ be the average speed of gas molecules in the x-direction.
When the block moves with speed $v_{0}$,the relative velocity of molecules hitting the front face is $(v + v_{0})$ and those hitting the back face is $(v - v_{0})$.
The number of molecules hitting the front face in time $\Delta t$ is $\frac{1}{2} n A (v + v_{0}) \Delta t$.
The momentum transferred per collision is $2m(v + v_{0})$.
The force on the front face is $F_{front} = \frac{1}{2} n A (v + v_{0}) \cdot 2m(v + v_{0}) = mnA(v + v_{0})^2$.
Similarly,the force on the back face is $F_{back} = mnA(v - v_{0})^2$.
The net drag force is $F = F_{front} - F_{back} = mnA[(v + v_{0})^2 - (v - v_{0})^2] = mnA(4vv_{0}) = 4(mn)Avv_{0}$.
Since $\rho = mn$,we have $F = 4\rho Avv_{0}$.
Using the kinetic theory,the average speed $v$ in one dimension is related to temperature by $\frac{1}{2}mv^2 = \frac{1}{2}kT$,so $v = \sqrt{\frac{kT}{m}}$.
Substituting $v$,the drag force is $F = 4\rho A v_{0} \sqrt{\frac{kT}{m}}$.