# Momentum and forces in gases

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1. May 7, 2015

### sgstudent

To obtain force from momentum we use the formula change in momentum/time. The time in the equation refers to the amount of time the force is exerted on the gas molecules.

So when considering the kinetic theory of gases we are taught that F ∝ change in momentum/ time so F ∝ 2mv/(2d/v) and hence F ∝ mv2/d. The thing that I don't understand is why the time is 2d/v. 2d/v gives us the time taken for the molecule to travel from one wall to the other. But the "time" in the change in momentum/time equation shouldn't be the time taken for the molecule to travel from one wall to another and instead should be the time the molecule spends in collision with the wall instead.

So why is the 2d/v always used?

2. May 7, 2015

### Staff: Mentor

This is the average rate of change of momentum for a single molecule. The instantaneous rate of change of momentum is zero during the time the molecule is traveling between the walls, and is very large for the brief time in which the molecule is in contact with the wall. If you add up all the average rates of change of momentum for all the molecules, you get the force exerted on a wall.

Chet

3. May 7, 2015

### sgstudent

Hmm but why would taking the time for the molecule to move from one wall to another give is the average force exerted? Shouldn't we had that very large value instead? I don't see how using 2d/v gives us the average force exerted on the wall by a single molecule..

Thanks

4. May 8, 2015

### Staff: Mentor

We want to add up the forces that all the molecules are exerting on the wall to get the total force, but they are not all hitting the wall as the same time. So this is the first step in beginning to take that into account.

Suppose I have a mass m, and I exert a force of 5 N on the mass for 10 seconds, and then I exert a force of 1 N on the mass for the next 50 seconds. What is the average force that I exerted on the mass over the entire 60 seconds? And, if I had exerted that same average force on the mass over the entire 60 seconds, would its momentum be the same as in the case of the two separate intervals?

Suppose I have a single molecule bouncing between two parallel walls, normal to the walls. The distance between the walls is d, and the velocity of the molecule when it hits each wall is v. So the molecule bounces of one of the walls every 2d/v seconds. Each time it bounces off the wall its velocity reverses, so that its change in momentum is 2mv. Assume that the time in contact with the wall is negligible compared to the transit time between successive strikes of each wall. So the time average of the force that each wall exerts on the molecule is mv2/d. This is also equal to the time average of the force that each molecule exerts on the wall. Now suppose you have N molecules bouncing back and forth, and they are all hitting the wall at different times. There are so many of them that, even though the contact time with the wall is very short, many of them are in contact with the wall at the same time (at different points during the contact interval). This has the effect of smoothing out the force at any one time for those in contact with the wall. To get the overall force acting on the wall at any one time, you just get Nnv2/d.

Chet