Kinetic theory of gases: rebound speed and force questions

[Amaterasu21]

Hi everyone,

I remember years ago at school memorising the derivation of the formula for pressure in the kinetic theory of gases, as explained in this Youtube video:


Thinking a little more deeply about this derivation there are two things I don't get:

1) At 0:53, the video says the molecule rebounds with the same velocity, so the change in momentum becomes -2mu.
Since momentum is conserved, that means the collision gives the wall a momentum of 2mu - it must move forward with some velocity when hit. That velocity will be tiny, since the wall is so much more massive than the molecule, but it won't be zero - the wall has to move forward a tiny amount.

Yet if the molecule bounced back with the same velocity, and its mass obviously doesn't change, that means the molecule's kinetic energy = 1/2mv^2 is unchanged. So all the kinetic energy in that collision stayed with the molecule, and none was transferred to the wall. So how can the wall move forward when hit?

I don't see how it's possible for the molecule to rebound with the same speed if both momentum and kinetic energy are conserved (and since this is an elastic collision, surely they have to be). It seems to me that the molecule must rebound at a slightly lower speed than it hit the wall at. And yet that can't be right, or else why don't molecules of air in a room all lose energy to the wall and slow down? I must be missing something here.

2) At 2:15, the implication is the molecule travels a distance of 2l in time t between collisions at speed u.

But surely for most of that time t, the molecule will be coasting between the walls and not colliding with them?

So how can that t be the same t in the equation F = dp/t? Surely THAT t is just the tiny amount of time the molecule spends in contact with the wall (and therefore exerting a force and changing the wall's momentum) rather than the whole time it takes to get to the other wall and back again?

If someone could clear up these two issues that'd be great! Thank you!

 

[sophiecentaur]

The relative magnitudes of velocities of molecule and container wall would be enormous (momentum conservation) - tending to infinity. The relative Kinetic Energy would actually mean that no energy would be lost to the wall.

But surely for most of that time t, the molecule will be coasting between the walls and not colliding with them?

The pressure in the cavity is a result of the rate of change of momentum against the sides. i.e. number of molecules hitting the side per second times momentum change of each molecule. The impacts are all due to different molecules and a rebounding molecule will hit another molecule in the gas, transferring momentum, eventually, to the other wall after a large number of later collisions. There is no time when molecules are striking the walls.

 

[anorlunda]

But surely for most of that time t, the molecule will be coasting between the walls and not colliding with them?

Not so. @Chestermiller said:

For a gas at atmospheric pressure, a typical value of the mean free path of a molecule (Wikipedia) is 68 nm. So there are lots of collisions of molecules within any practical volume of gas, even as ideal gas behavior is approached. This is how pressure variations are transmitted through a gas is cases where gas is accelerating or where the gas is in a gravitational field (which, according to equivalence principle, is equivalent to acceleration).

Think about it. Without frequent collisions, a gas could not conduct sound.