Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.17 m on a side. What is the average force the molecule exerts on one of the walls of the container?

The last part of the question has some details missing which is ; (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) molecule v=482 m/s molecule momentum=2.56 x 10^(-23)

Explanation:

The momentum of the molecule is 2.56 x 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.time required = distance /speed= 0.17 X 2/(482 m/s)Average force = impulse / time= 2 x 482 x 2.56 x 10^(-23) / (0.17 x 2)= 7.76 x 10^20N, is the average force the molecule exerts on one of the walls of the container.

Answer8.0m;