Gas Pressure

1. Nov 14, 2004

jamie_o

Hello, I am having some difficulty following the method for finding an equation for an ideal gas. There are a few different forms, but I'm proving 1. For an indiviual particle of a gas in a cube container side length L, it is travelling with a velocity of u1 on the x axis (its x component of velocity is u1). This collides with one surface and rebounds elastically. This would have a velocity -u1 as no kinetic energy was lost. I want to find the force this particle exerts on the wall by its x component. So to find Force i would use rate of change of momentum. the distance before it collides with the surface is 2L. so time = 2L/u1.

so for my equation I now have force = change of momentum
2L/u1

change in momentum equals mv - mu. according to the principal of conservation of momentum would i be correct in saying mv - mu = 0 ?? (v = final velocity, u = initial)

if so, -mu1 - mu1 is the change in momentum -2mu1. Now i was told the change in momentum should be equal to 2mu1. Which leaves me with a problem. Does -2mu1 = 0 ??? if so can i easily get 2mu1 by adding it to both sides? which gives me a positive change in momentum. Seems odd to me, can someone please explain that and if that would be the correct way of doing it? I have no problems getting the P = 1/3 rho (mean velocity squared) equation from that, but the change in momentum is bothering me. I could just remember 2mu1, but I would rather know how to work it out properly. Any help is very much appreciated, thanks

2. Nov 14, 2004

Diane_

Another term for "change in momentum" is "impulse". You can get the impulse by taking the product of the force exerted and the time it is exerted. Now you need to find a way to get rid of the time: consider that the pressure is the force exerted over an area, and this will depend on the force exerted by each particle (which you're working on), but also on the total number of particles that strike that area in a given time.

Is that enough?