Momentum and the pressure of an ideal gas. Easy Question.

[jmcgraw]

My book (Halliday, 6th ed: Section 20-4), uses the momentum of the individual molecules in a gas to derive the pressure of the gas. They imagine the molecules hitting a wall. I'm a little rusty on my memory of conservation of momentum, so this equation is confusing me a bit:

(delta)p_x = (-mv_x) - (mv_x) = -2mv_x

They then say that the molecule delivers +2mv_x of momentum to the wall.

Are they approximating the wall as much more massive than the molecule, so the molecule's speed is unchanged (velocity opposite)? So to get from +mv_x to -mv_x you would take

mv_x -mv_x -mv_x = -mv_x

to get the change in momentum?

I'm thinking I'm probably right, but I just feel a little queezy about it. Like there is something I'm not getting. I guess what I don't understand is the sentence: "the momentum (delta)p_x delivered to the wall by the molecule during the collision is +2mv_x."

Could someone possibly start from scratch and show how all this comes from the conservation of momentum?

I hope I made sense.

Thanks a lot.

 

[Pyrrhus]

You know what is an elastic collision right?, if a particle with speed v crashes against a perfectly elastic wall, it will go the opposite direction with the same speed v, so the particle will have a change in momentum.

\Delta p = -mv_{x} - mv_{x}

 

[lightgrav]

Delta p _OF_ the gas molecule is p_final - p_initial .
the original momentum was (apparently) mv_x .

Thinking of the center-of-momentum reference frame,
a perfectly elastic collision has relative speed afterward
being equal to the relative speed before collision.

Yes, the wall is being thought of as one object,
not as a bunch of individual molecules.
So the wall does not move after the collision.
(this is more general than it sounds; right wall
is connected to the left wall, pushed leftward.)

This means that the total momentum is constant,
so the momentum change of the WALL
is the negative momentum change of the molecule.