Suppose an oxygen molecule traveling at this speed bounces b

Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.16 m on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule\'s velocity is perpendicular to the two sides that it strikes.) molecule v=482 m/s molecule momentum=2.56*10^(-23)

Solution

The momentum of the molecule is =2.56*10^(-23) . Particle hits the wall and bounces. Momentum is reversed. Change in momentum = impulse . This is Force x time. Momentum change happens at a wall after each trip. particle goes to the other wall and comes back.

time required is =distance /speed

= .16 X 2/(482 m/s)

Avarage force= impulse /time

=2*482*2.56*10^(-23)/(.16*2)

=7.7*10^(-20) N