How Is Time Factored Into the Derivation of the Ideal Gas Law?

[eddywalrus]

Here is a screenshot from a page from a textbook that explains how to derive the ideal gas law:

In the third bold line, I don't understand how "time" in force = (change of momentum)/(time) is equal to 2x/u (the time it takes for the particle to travel to the opposite face and back again) -- I always assumed that:

impulse = Force x time
change in momentum = Force x time
where time in this case refers to the time of contact between the two colliding objects? Furthermore, since the particle doesn't change its momentum over the duration of traveling to the opposite face and back again (but instead changes momentum during its collision with the container wall), shouldn't the "time" in this case refer to the time of collision?

Thank you so much for all your help!

 

[Evanish]

I'm pretty sure you can derive (change of momentum)/(time) from some basic formulas.

F = ma
a = Δv/Δt
Δp = mΔv

F = ma and a = Δv/Δt gets you F = mΔv/Δt

F = mΔv/Δt and Δp = mΔv gets you F = Δp/Δt

 

[eddywalrus]

I'm pretty sure you can derive (change of momentum)/(time) from some basic formulas.

F = ma
a = Δv/Δt
Δp = mΔv

F = ma and a = Δv/Δt gets you F = mΔv/Δt

F = mΔv/Δt and Δp = mΔv gets you F = Δp/Δt

Thank you for your help, but I think you misunderstood my question -- I probably should have made it clearer. My bad, sorry.

I get how you would derive force = (change in momentum)/(time), but I'm unsure of why "time" in this instance is the time it takes for the particle to travel to the other face and back instead of the time of collision or contact between the particle and the container wall.

Thank you!

 

[256bits]

the time of collision or contact between the particle and the container wall

Already considered in the statement,
+mu - (-mu) = 2mu
for the particle interacting with the wall.

time it takes for the particle to travel to the other face and back

As in the textbook,
time between collisions = distance /speed = 2x/u

The particle interacts with the wall only once every interval, and not continuously during the interval.
So we want to find a force, that if acting continuously, would give the same force on the wall as from the intermittent collisions of the particle with the wall.

 

[eddywalrus]

Already considered in the statement,
+mu - (-mu) = 2mu
for the particle interacting with the wall.As in the textbook,
time between collisions = distance /speed = 2x/u

The particle interacts with the wall only once every interval, and not continuously during the interval.
So we want to find a force, that if acting continuously, would give the same force on the wall as from the intermittent collisions of the particle with the wall.

Thank you very much for your explanation -- I understand it now!