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  • Consider a rectangular block of wood moving with a velocity v_0 in gas at temperature T and mass density rho

Consider a rectangular block of wood moving with a velocity v0 in 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 Av_{0}\sqrt{\frac{KT}{m}}, where m is the mass of the gas molecule.

Answers (1)

We assume ‘p’ to be the number of molecules per unit volume. Hence ‘p’ is the per unit volume molecular density. Let v be the velocity of gas molecules.

The molecules of the gas strike the front face and the back face of the box when it moves. The back face relative velocity = (v-v_0)

Change in momentum on the front face of the box = 2m(v+v_{0}) and the change in momentum on the back face of the box = 2m(v-v _0).

Nf =Total number of molecules striking the box’s front face = \frac{1}{2}[A.(v+v_{0}) \Delta t] p

Nf = \frac{1}{2}(v+v_{0}) A.p. \Delta t

Nb = Total number of molecules striking the box’s back face

 =\frac{1}{2}[A.(v+v_{0}) \Delta t] p

Nb

=\frac{1}{2}(v+v_{0}) A.p. \Delta t

Total change in momentum at the front face can be calculated s below:

Pf = 2m(v+v_{0}) \times Nf = 2m(v+v_{0}) \times \frac{1}{2}(v+v_{o}) A.p. \Delta t

Pf = - m(v+v_0)2 A.p. \Delta t in the backward direction.

Force on front end = Ff = Pf = - m(v+v_0)2 A. p in the backward direction

In the same way, the force on the back face, Fb = + m(v-v_{0})2 A. p

So, the net force can be calculated as:

- m(v+v_{0})^{2} A. p+ m(v-v_{0})^{2} A. p

It can be expanded and simplified as,

= -m A p 4v.v_{0}

The magnitude of the dragging force= 4m v.v _{0}.A.p

Kinetic energy for a molecule of gas can be calculated as:

KE = \frac{1}{2}mv^{2} = \frac{3}{2} Kb T

Hence, v = \sqrt{\frac{Kb T}{m}}  

So, we can write the drag force as 4m.A. p. v_o \sqrt{\frac{Kb T}{m}}

Posted by

infoexpert24

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