question_answer 45)

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

Answer:

                  Let \[N=\] number of molecules per unit volume. Velocity of molecules of gas w.r.t. the block \[=(\upsilon +{{\upsilon }_{0}})\]. Average number of collision of gas molecules with the block in time \[\Delta t=\frac{N(\upsilon +{{\upsilon }_{0}})A\,\Delta t}{2},\] where A is the area of the face of the block.                 If \[m\] is the mass of the gas molecule, then momentum transferred to the block per collision \[=2m(\upsilon +{{\upsilon }_{0}})\]             Net force or drag force, \[F=mnA\]                 \[[{{(\upsilon +{{\upsilon }_{0}})}^{2}}-{{(\upsilon -{{\upsilon }_{0}})}^{2}}]=4m\,\,nA\,\,\upsilon \,\,{{\upsilon }_{0}}\]                 Now \[mn=\rho \] (density of gas)                 \[\therefore \]\[F=4\,\,\rho \,A\,\upsilon \,{{\upsilon }_{0}}\]                 Now \[\frac{1}{2}m{{\upsilon }^{2}}=\frac{1}{2}kT\] or \[\upsilon =\sqrt{\frac{kT}{m}}\]             \[\therefore \] \[F=4\,\rho \,A\,{{\upsilon }_{0}}=\sqrt{\frac{kT}{m}}\]