, where m is the mass of the gas molecule.
Answer:
Let ρm represent the number of molecules per unit volume
Then the expression for the change in momentum by a molecule on front side is
= 2m (v + v0)
Similarly, the expression for change in momentum by a molecule on backside is
= 2m (v – v0)
A number of molecules striking the front side is given as follows:
= 1/2 [A(v+v0)∆t] ρm
The number of molecules striking the backside is
=1/2 [A(v-v0)∆t] ρm
Upon solving the above equation by considering the KE of the gas molecule,
we get the dragging force as follows:
= 4m A ρmv0√kgT/m