Question

Consider a rectangular block of wood moving with a velocity v₀ in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v₀ is A. Show that the drag force on the block is 4ρAv₀√(kT/m), where m is the mass of the gas molecule.

Answer 1

n = no. of molecules per unit volume

v_rms = rms speed of gas molecules

When block is moving with speed v_o, relative speed of molecules w.r.t.

front face = v + v_o

Coming head on, momentum transferred to block per collission

= 2 m (v + v_o), where m = mass of molecule.

No. of collission in time Δt = 1/2(v + v_o)nΔtA, where A = area of cross section of block and factor of 1/2 appears due to particles moving towards block.

Momentum transferred in time Δt = m(v + v_o)²AΔt  from front surface

Similarly momentum transferred in time Δt = m(v -v_o)²nAΔt  from back surface

Net force (drag force) = mnA [(v + v_o) - (v - v_o)²] from front

= mnA (4vv_o) = (4mnAv)v_o)

= (4ρAv)v_o 

We also have (1/2)mv² = (1/2)kT (v - is the velocity along c-axis)

Therefore, v = √kT/m

Thus drag = 4ρA√(kT/m)v_o