Why is time interval 2L/V in kinetic theory collision formula?

[terryds]

F = Δp/Δt

http://www.sumoware.com/images/temp/xzhhjttxobbodxgr.png
(Sorry for bad sketch.. hehe)

In the kinetic-molecular theory of gas, the Δp is -2mv (since it's supposed to be elastic collision)
But, why the Δt is 2L/V ??
I think it must be a very small time (since the impact is done in a very very small time)
Why the time is the interval time before the velocity gets back to the initial velocity ?

 

[DrClaude]

You want the average, so the actual duration of the collision is not the relevant time scale. 2L/v is the time it will take for a molecule to travel to the other wall and come back to hit again. Therefore, on a given wall a molecule will have a change of momentum of Δp every 2L/v, on average.

 

[PWiz]

It should be $Δ t=\frac{2L}{V_x}$. V is the velocity in 3D, and the distance covered may not necessarily be L (remember that $<V^2>=<{V_x}^2>+<{V_y}^2>+<{V_z}^2>$, and the velocity components in the $y$ and $z$ direction may be non-zero). The particle starts at one wall, travels a distance of $L$ due to it $V_x$ velocity component, collides, and returns along its original trajectory to the wall covering a distance of L again (total distance traveled is $2L$). Therefore, $Δt=\frac{2L}{V_x}$. On a side note, to avoid confusion I recommend that you use $v$ to denote the velocity instead, as $V$ is almost always used to denote volume in these kind of questions.

 

[terryds]

It should be $Δ t=\frac{2L}{V_x}$. V is the velocity in 3D, and the distance covered may not necessarily be L (remember that $<V^2>=<{V_x}^2>+<{V_y}^2>+<{V_z}^2>$, and the velocity components in the $y$ and $z$ direction may be non-zero). The particle starts at one wall, travels a distance of $L$ due to it $V_x$ velocity component, collides, and returns along its original trajectory to the wall covering a distance of L again (total distance traveled is $2L$). Therefore, $Δt=\frac{2L}{V_x}$. On a side note, to avoid confusion I recommend that you use $v$ to denote the velocity instead, as $V$ is almost always used to denote volume in these kind of questions.

So, when it has non-zero velocities in x,y, and z directions, Δt is no longer 2L/v , right ?

 

[DrClaude]

So, when it has non-zero velocities in x,y, and z directions, Δt is no longer 2L/v , right ?

Wrong. You can decompose the motion along x, y and z, and look only along one axis. As PWiz correctly clarified, you have $\Delta t = 2L/v_x$: you need to consider the component of velocity along one direction.

 

[PWiz]

So, when it has non-zero velocities in x,y, and z directions, Δt is no longer 2L/v , right ?

Just remember that time over here is always equal to (2×distance traveled )/(velocity in the direction of travel). If for example you take the distance only along the y axis, then you must use the $v_y$ component. If you take the distance traveled in the xy plane, then use the velocity in the xy plane.

 

[terryds]

Wrong. You can decompose the motion along x, y and z, and look only along one axis. As PWiz correctly clarified, you have $\Delta t = 2L/v_x$: you need to consider the component of velocity along one direction.

Hmm..
I imagine if v_z is much greater than v_x
So, I think the gas molecule will collide before it travels 2L_x..

 

[terryds]

Just remember that time over here is always equal to (2×distance traveled )/(velocity in the direction of travel). If for example you take the distance only along the y axis, then you must use the $v_y$ component. If you take the distance traveled in the xy plane, then use the velocity in the xy plane.

Hmm.. I see in ,
Can you please explain me why v_x = v_y = v[sub]z[/sub] ( Underline means that the average of velocity, I know it should be an over bar, but I don't know how to Latex haha)

 

[DrClaude]

Hmm..
I imagine if v_z is much greater than v_x
So, I think the gas molecule will collide before it travels 2L_x..

But it will not change how long it will take before it hits again the same wall which lies in the yz plane.

I made a drawing: the green molecule will take the same time to come back to that wall regardless of whether it takes the blue or the red path, provided that $v_x$ is the same.

 

[DrClaude]

Can you please explain me why v_x = v_y = v[sub]z[/sub] ( Underline means that the average of velocity, I know it should be an over bar, but I don't know how to Latex haha)

Because of collisions between molecules, at equilibrium there is no preferred direction, and the average velocity should be the same along any axis.

 

[PWiz]

Hmm.. I see in ,
Can you please explain me why v_x = v_y = v[sub]z[/sub] ( Underline means that the average of velocity, I know it should be an over bar, but I don't know how to Latex haha)

To use latex, just place the variables between two pound signs (i.e. #). When you're typing a post, you will see that near the post button, there is a latex preview button and a latex guide (I don't remember the exact position because I usually post through my phone).

The principal assumption in the kinetic theory of gases is that the number of gas particles is so large that we can say that the average components of the velocity of each particle will be approximately the same. One particle might have an unusually large $v_z$ component, but when you'll take the average over a very large number, they will all have the same average velocity. (Technically we are comparing the average of the sqaured velocity terms - hence we get a formula for the root mean square velocity, not the average velocity.)