Why is 2L used as the distance in the derivation of 1/3nMc^2 = nRT?

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SUMMARY

The discussion clarifies the use of 2L in the derivation of the equation 1/3nMc² = nRT, where L represents the distance between the walls of a container. The time interval, t, is defined as t = 2L/v, accounting for the round trip of a molecule colliding with the wall. This approach calculates the average force exerted by numerous molecules in random motion, emphasizing the importance of averaging over many identical events rather than focusing on individual collisions. The concept of average force is essential in kinetic theory, where it reflects the collective behavior of gas molecules.

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sgstudent
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In the derivation the first step used F=Δmv/t and for t, they used t=2L/v where L is the distance between one end to the other end of the wall.

But I don't understand why we use 2L as the distance. Isn't the force exerted by that molecule only for the very short period where the molecule is in contact with the wall only? For example when we look at a car crash we look at that short moment when the car comes to a stop from an extremely high speed- shouldn't the same be applied?
 
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sgstudent said:
In the derivation the first step used F=Δmv/t and for t, they used t=2L/v where L is the distance between one end to the other end of the wall.

But I don't understand why we use 2L as the distance

actually they calculate average force and the time interval is between two collisions taken as a round trip by a molecule after hitting a wall.
and there is a factor 2 also for the momentum change
F = delta p = 2 m v(x) x delta t and delta t= 2L/v(x)
 
drvrm said:
actually they calculate average force and the time interval is between two collisions taken as a round trip by a molecule after hitting a wall.
and there is a factor 2 also for the momentum change
F = delta p = 2 m v(x) x delta t and delta t= 2L/v(x)
Could you explain why using the change in momentum from one end to the other for the duration need (t) would give us the average force? I have not been exposed to the concept of an average force before so I'm not sure how to use it. Thanks!
 
Last edited:
sgstudent said:
Could you explain why using the change in momentum from one end to the other for the duration need (t) would give us the average force? I have not been exposed to the concept of an average force before so I'm not sure how to use it. Thanks!

i think the term 'average' is used by people if there are variations in 'individual' events but sum /total of the effect carried out can be averaged over a large number of identical measurements/events. due to large number of molecules in random motion inside the container and colliding with each other as well as with the walls of the container ,an observer can think of averaging over the state of motion rather than takinng individual molecules and adding individual characteristic path , motion and impulse transferred to the wall of the enclosure.

in kinetic theory the picture is almost above therefore the average force/average velocity and other terms are being used'
one can visit the following to
get a clear picture;

http://galileoandeinstein.physics.v...dfs/10_1425_web_Lec_31_KineticTheoryGases.pdf
 

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