Consider a rectangular block of wood moving with a velocity v0 in a gas of mass  density  ρ and at temperature T. Assuming the velocity is along x-axis and the  area of cross-section of the block perpendicular to v0  is A.The drag force acting  on the block is  NρAv0kTm, where m is mass of each molecule. Find N.

Using kinetic theory of ideal gases,n = number density of moleculesm = mass of each gas moleculevRMS = RMS velocity of each gas moleculev = velocity of molecule that is moving along x-axisAs RMS velocity along each direction is equal  vRMSx=vRMSY=vRMSZ=vNow,  vRMS2=vRMSx2=vRMSy2=vRMSz2=3v2⇒v2=vRMS23      Let block move towards right. It faces “front” molecules that collide with the block  while moving towards left. Also there are “back” molecules that collide with the  block from behind while moving towards right.Relative speed of “front” molecules before collision =  v+v0Relative speed of “back” molecules before collision =  v−v0Since, collisions are elastic, relative speed of molecules remains same before and  after collision. Since, block is  heavy, so it does not change its speed and direction  of motion.After collision, “front” molecules move towards right, and “back” molecules move  towards left.Let  v1 and v2  be speeds of “front” and “back” molecules along x-axis respectively  after collision.So,  v+v0=v1−v0⇒v1=v+2v0And,  v−v0=v0−−v2⇒v2=v−2v0Magnitude of momentum transferred by each “front” molecule to the block = mass  of  moleculechange   in  its  velocity=mv+2v0−−v=2mv+v0 Its direction is towards left.Magnitude of momentum transferred by each “back” molecule to the block =mass  of  moleculechange   in  its  velocity=m−v−2v0−v=2mv−v0 Its direction is towards  right.Consider an interval of time Δt. All “front” molecules within a distance (along x-axis) v+v0Δt will collide with the block. And “back” molecules within a distance (along  x-axis) v−v0Δt will collide with the block.Momentum transferred by “front” molecules in time  Δt = half  the  number  of  molecules  in volume  Av+v0Δtmomentum  transferred  by  each  "front"  molecule12n Av+v0Δt2mv+v0=nmAv+v02Δt   (The factor of half is considered because half of molecules move towards the block  and other half towards the block, in each direction).Similarly, momentum transferred by “back” molecules in time  Δt = half  the  number  of  molecules  in volume  Av−v0Δtmomentum  transferred  by  each  "back"  molecule12n Av−v0Δt2mv−v0=nmAv−v02Δt   Total momentum transferred to the block in time Δt is =  nmAv+v02Δt−nmAv−v02Δt=  4nmAvv0ΔtHence, drag force on the block =  momentum  transferred  to  block  time=4nmAvv0But  mn=ρ (density of gas)And  vRMS=3kTmSo,  v2=vRMS23=3kTm3=kTmHence, drag force = 4ρAv0kTm .