Some Kinetic Gas Theory questions

  • Thread starter carvas
  • Start date
  • #1
6
0
Hi there,

I'm having a problem interpreting how a velocity interval defined by [itex]dv^3 = dv_x dv_y dv_z[/itex], being an isotropic case, why do we write it like this:

[itex] dv^3 = 4 \pi v^2 dv [/itex]

And also, how can I derive the molecular impingement rate over a surface? I saw at a book, but I don't understand it, that, considering only particles with velocity [itex] v_x [/itex] will hit the surface, in order to get the molecular impingement rate, [itex] J [/itex], we have to:

[itex] J = \frac{1}{V} \int_ 0^{\infty} v_x dN = \frac{n\, v_{avg}}{4} [/itex]

Where [itex] dN [/itex] is the number of molecules, [itex] n = N/V [/itex], being [itex] V [/itex] the volume and [itex] v_{avg} [/itex] the average velocity.

Thank you!
 
Last edited:

Answers and Replies

  • #2
395
42
The rewriting is due to using spherical coordinates (|v|,phi,theta) instead of cartesian coordiantes (vx,vy,vz).
Spherical coordinates are indicated if your integrand depends only on the radial component |v|, not on the direction. The reason that the integrals over theta and phi are not seen is that they are already carried out (which for a given |v| gives 4pi |v|²). I cannot help you with the 2nd part at the moment.
 
Last edited:
  • #3
Ken G
Gold Member
4,438
333
If the direction x is perpendicular to the surface in question, then v_x carries particles across the surface. The rate they are carried across is the speed times the particle number density-- note the units of that is a per area per time. So to get a number of particles crossing, you need to multiply that rate by the area and the time. It should make perfect sense that the number of particles should be proportional to the area and the time, and it should also be proportional to the speed and the density of particles. There is nothing else it should depend on, so multiplying them all together should work, and indeed it does give the unit of "particles" when you do that.
 
  • #4
6
0
The rewriting is due to using spherical coordinates (|v|,phi,theta) instead of cartesian coordiantes (vx,vy,vz).
Spherical coordinates are indicated if your integrand depends only on the radial component |v|, not on the direction. The reason that the integrals over theta and phi are not seen is that they are already carried out (which for a given |v| gives 4pi |v|²). I cannot help you with the 2nd part at the moment.
Thank you Timo!

So you have 3 coordinates in cartesian space [itex] (v_x, v_y, v_z) [/itex], and you have that [itex] v^2 = v_x^2 + v_y^2 + v_z^2 [/itex], so you're saying, that in analogy with
position in spherical coordinates like [itex] (r,\theta,\phi) [/itex] I can just make the same thing to velocities [itex] (v,\theta,\phi) [/itex], right?

Just another thing, the differential element for the velocities is just: [itex] dv^3 = v^2\,dv\,\sin \theta\,d\theta\,d \phi [/itex] ?
 
Last edited:
  • #5
6
0
If the direction x is perpendicular to the surface in question, then v_x carries particles across the surface. The rate they are carried across is the speed times the particle number density-- note the units of that is a per area per time. So to get a number of particles crossing, you need to multiply that rate by the area and the time. It should make perfect sense that the number of particles should be proportional to the area and the time, and it should also be proportional to the speed and the density of particles. There is nothing else it should depend on, so multiplying them all together should work, and indeed it does give the unit of "particles" when you do that.
Thank you Ken!

Yes, making a dimensional analysis really makes it easy to understand. :)
 
  • #6
395
42
So you have 3 coordinates in cartesian space [itex] (v_x, v_y, v_z) [/itex], and you have that [itex] v^2 = v_x^2 + v_y^2 + v_z^2 [/itex], so you're saying, that in analogy with
position in spherical coordinates like [itex] (r,\theta,\phi) [/itex] I can just make the same thing to velocities [itex] (v,\theta,\phi) [/itex], right?

Just another thing, the differential element for the velocities is just: [itex] dv^3 = v^2\,dv\,\sin \theta\,d\theta\,d \phi [/itex] ?
"Yes" to both. Except perhaps for the ambiguity of your last equation where "v" appears on both sides of the equation but means different things - I think we both know what is meant.
 
  • #7
411
34
And also, how can I derive the molecular impingement rate over a surface?

Thank you!
The molecular flux rate (∫ = nivi)for an ideal gas is the product of the number of molecules per cubic meter having a component of motion toward the incident surface (ni) and the mean axial molecular velocity normal to and toward that surface (vi).

Under conditions of equilibrium, ni = n/2, where n is the molecular number density in molecules per cubic meter: n = P/kT.

vi = (2/π)1/2 σ, where σ is the root-mean-square axial velocity (normal to and toward the surface): σ = (P/nm)1/2, where m is the molecular mass of a gas having only a single molecular mass, and the mean impulse mass of a gas having more than one molecular mass.
 

Related Threads on Some Kinetic Gas Theory questions

Replies
4
Views
2K
Replies
0
Views
2K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
580
  • Last Post
Replies
1
Views
827
  • Last Post
Replies
4
Views
1K
Replies
1
Views
7K
Replies
13
Views
1K
  • Last Post
Replies
10
Views
2K
Top