Question

In a thought experiment, a cubic box of side length 0.01 mm contains three particles. One particle moves just in the x-direction, colliding elastically with the left and right walls of the cube. The second atom moves just in the y-direction, colliding elastically with the front and back walls of the cube. The third atom moves just in the z-direction, colliding elastically with the top and bottom walls of the cube. All three atoms have speed 500 m/s and mass 6.7x10-27 kg. What is the average pressure in the box?

A. 1.7 × 10-3 Pa

B. 1.7 × 10-6 Pa

C. 3.3 × 10-3 Pa

D. 5.0 × 10-6 Pa

E. 1.01 × 10-5 Pa

F. 3.10 x 10-5 Pa

Answer 1

Pressure exerted by each particle on walls perpendicular to it's velocity, are equal and is average pressure inside box. Let's consider particle moving along x axis exerting force on walls parallel to y-z plane. Consider collisions of this particle with right side wall. Every time it collides elastically with wall, it's momentum changes by 2mv ( there is change in direction of velocity after collision). m is mass and v is speed of particle. It collides with wall at regular intervals of 2L/v, L is length of cube. Hence number of collision per second with right wall = v/2L.

Hence change in momentum of particle per sec =

(2mv)(v/2L) = mv^2/L

By Newton's second law, this is average force acting on particle. By Newton's third law, this is average force acting on wall.

Average pressure acting on wall or inside cube

= force / area = mv^2/L^3

= 6.7×10^-27 kg*(500m/s)^2 / (10^-5m)^3

= 1.7×10^-6 Pa