Problem 1. Pressure of a gas: First consider one atom of mass m bouncing elastic

ID: 1704063 • Letter: P

Question

Problem 1. Pressure of a gas: First consider one atom of mass m bouncing elastically
inside a perfect cube of interior size L L L = V (volume V ). The atom bounces
elastically of the interior walls of the cube. Assume the walls have innite mass and are oriented perpendicular to the x, y and z axes. When the atom bounces, the impulse is normal to the wall that it bounces o of. The initial velocity of the atom is v = (vx; vy; vz).

(a) What is the time-averaged force that this atom exerts on a wall whose surface is
perpendicular to the x axis?
Now assume that there are N >>1 such atoms inside the cube. The atoms may have
different masses m subi. The directions vi of their velocities are random and equally likely to point in any direction on the unit sphere. The total kinetic energy of these atoms is U = (3/2)NkBT. The atoms do not interact with each other, they only bounce elastically of of the walls.

(b) Now what is the pressure of this gas as measured by the time-averaged force per unit area on the walls? Show that it obeys the ideal gas law.

Explanation / Answer

The mass of the atom is m

The length of the cube is L and the volume V = L3

The velocity of the particle is v = (vx,vy,vz)

The surface which is perpendicular to the X axis will be parellel to YZ plane

As there are no collissions across the particles,

The change in momentum of an atom moving along X axis with the speed vx is

P = m*vx - m(-vx)

P = 2mvx

This gives the impulse acting on the wall by one such atom

J = P = 2mvx

From Newton's second law of motion

J = F*t

The average force experienced by the wall of the container is

F = 2mvx/t

Therefore the time-averaged force that this atom exerts on a wall whose surface is
perpendicular to the x axis is

F = 2mvx/t

Then the number of atoms moving along X direction through the area A by a small segment of length x is given by:

N = 1/2V * A* x

N = 1/2L3 * L2* vx*t

N/t = vx/2L

Therefore the average force on the walls is

F = 2mvx( vx/2L)

F = mvx2/L

In a similar methedology we can derive the forces on the remaining faces also.

So for the average of all these atoms moving at random directions we consider the Root Mean Square speed of the atom

vRMS = [(vx)2 +(vy)2 +(vz)2]

From the symmetry

vx2= 1/3 * vRMS2

The total time averaged force on the wall is given by

F = mvRMS2/3L

So the pressure of the gas in the container is

P = F/A

P = mvRMS2/3L*L2

P = mvRMS2/3L3

P = 2/3L3 *1/2 *mvRMS2

P = 2/3 * U/V (V =L3 )

Where U is the total kinetic energy given by

U = 3/2(NkBT)

PV = 2/3 *3/2(NkBT)

PV = NkBT

Which represents the ideal gas law with Boltzman's constant kB

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Problem 1. Pressure of a gas: First consider one atom of mass m bouncing elastic

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