Kinetic theory

Question:
1. A square tray of sides 0.10 m is placed on a table. In the tray are 500 ball bearings, each of mass 1.0 X 10^-6 kg. When the tray is shaken, the ball bearings move with a root-mean-square speed of 2.0 x 10^-2 ms^-1 and collide with the walls of the tray. If the collisions are elastic, calculate the force on each wall of the tray due to the collisions of the ball bearings.

2. An experiment is carried out inside a vacuum system where the pressure is 1.0x10^-7 Pa and the temperature 20 ^oC.
Estimate the number of gas molecules per cubic meter in the system.

Answers:

1. We will assume that on average each side of the tray receives impacts due to one quarter of the balls. In other words their velocities are distributed uniformly, 125 balls hit each wall of the tray.

At each impact the change of momentum of a ball is given by:
mv = 1.0 X 10^-6 x 2.0 x 10^-2 = 2.0x10^-8 Ns
Between impacts each ball moves 0.2 m (there and back across the tray). Therefore the number of impacts per second on the face due to an individual ball is 0.2/2x10^-2 = 10
Therefore due to one ball the change of momentum per second at the wall of the tray is:

Change of momentum per second = Number of impacts x momentum change due to one impact = 10 x 2.0x10^-8 = 2x10^-7 Ns

But the rate of change of momentum is the force and so:
the force due to one ball on the wall is 2x10^-7 N

But on average 125 balls hit this wall and so the total force on this wall is 125x2x10^-7 = 2.5x10^-5 N

2. Pressure (P) = 1.0x10^-7 Pa Temperature (T) = 20 ^oC = 293 K
(We will need Boltzmann's constant for this one)

If the gas contains n moles and N molecules in a volume V then: PV = nRT = NkT where k is Boltzmann's constant [1.38x10^-23 JK^-1]
N/V = P/kT = 1.0x10^-7/[1.38x10^{- 23}x293] = 2.47x10¹³ molecules per cubic metre.

N.B – note the use of K rather than degrees C in this question.