# Kinetic theory

1. Dec 15, 2012

### Deathnote777

Hey guys! I find it confusing to prove "mv^2/L", the average force acted by the air molecules on a wall in a box. My textbook tells me an assumption "the time during each molecules colliding with the wall of the container is negligible compared with the time elapsed between successive collision" And my textbook tells me the change in momentum = 2mv (I agree with it), the delta t = 2L/v where L is the length of a box <--- I dun understand this one. Why the time of impact is this ? How is it related to the assumption ? I dun understand

2. Dec 15, 2012

### Khashishi

First consider a box with just one molecule in it, and the molecule bounces back and forth between two walls with the back and forth component of velocity = v_x. It takes two passes of the box length between collisions with the far wall. So the time is just 2L/v_x.

When there are more molecules, there are collisions between molecules. But the average collision rate with the wall per molecule is still the same. I'm not sure how to prove this, but maybe you can picture it this way: with 2 molecules, they can collide in the center of the box. So a molecule going on its return trip to the wall could be deflected away and take much longer. But, by conservation of momentum, the other molecule must be knocked toward the wall, so it takes shorter to hit the wall. It all works out so the average is the same.

3. Dec 15, 2012

### Deathnote777

I dun understand why the time "2L/Vx" is taken, the time should be time of impact instead of this one. The time of impact should be much smaller

4. Dec 15, 2012

### Staff: Mentor

Go back to the single particle moving with speed V in the plus/minus x direction as it bounces between the two walls. The particle starts moving left, hits the left-hand wall a time 0, and rebounds to the right.

When does the particle hit the right-hand wall? At time zero it had just started moving right from the left wall, at speed V, and it has to travel the width of the box to get to the right-hand wall. So it will hit the right-hand wall at time L/V, and then rebound to the left at speed V.

When will it get back to the left-hand wall? It has to travel the width of the box, back again from the right-hand wall, and it's moving at speed V, so again the travel time will be L/V. And it started at the right-hand wall at time L/V, so will be back at the left-hand wall at time 2L/V.

Thus the particle hits the left-hand wall at times 0, 2L/V, 4L/V, ....; and hits the right-hand wall at times L/V, 3L/V, 5L/V, .... For both, it's one momentum change of magnitude 2mV every 2L/V seconds.

And because force is defined as the change in momentum per unit time, getting a kick of magnitude 2mV every 2L/V seconds means that the force is
$\frac{2mV}{2L/V} = \frac{2mV^2}{L}$

5. Dec 15, 2012

### Deathnote777

I understand what you say. But why the time when the molecules are travelling in the air is taken into account ? It shouldn't. Again, the time should be the IMPACT TIME, the time when the air molecules are contacting/colliding with the wall. E.g. When i throw a ball to the wall with speed v, it rebounces to me after 5 seconds at speed v. The average force will NOT be 2mv/5. Instead, it should be like 2mv/0.2 where 0.2 is the time when the ball is colliding with the wall.

6. Dec 16, 2012

### Staff: Mentor

To compute the average force over a period of 5 seconds, I have to include in the average not only the .2 seconds when the force is 2mv/.2 but also the 4.8 seconds when the force is zero.

Another way of thinking about it: if I hit something twice a second, I'll be applying twice the force over time than if I were hitting it once a second - even though the force of each blow is the same.

7. Dec 16, 2012

### Deathnote777

Thx. I think I get it. I shouldn't consider only the time interval when molecules are colliding with the wall. Right ? Then what's about the assumption "the time during each molecules colliding with the wall of the container is negligible compared with the time elapsed between successive collision" ?

8. Dec 16, 2012

### sophiecentaur

This is not the time for an impact, it is the time between impacts. Considering the change of momentum during impact removes the need to be concerned with how long the impact takes.