Question

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?

Answer 1

Answer: 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.