Some Kinetic Gas Theory questions

In summary: P/m) = [n/π] √(P/m) = [n/(πm)]√P. In summary, the molecular impingement rate over a surface can be derived by taking the product of the number of molecules per cubic meter with a component of motion toward the surface and the mean axial molecular velocity normal to and toward that surface, under conditions of equilibrium.
  • #1
carvas
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!
 
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  • #2
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.
 
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  • #3
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
Timo said:
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] ?
 
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  • #5
Ken G said:
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
carvas said:
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
carvas said:
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.
 

FAQ: Some Kinetic Gas Theory questions

What is the kinetic theory of gases?

The kinetic theory of gases is a scientific model that describes the behavior of gases based on the motion of their particles. It states that gas particles are in constant, random motion and that their collisions with each other and with the walls of their container cause pressure, temperature, and other properties of gases.

How does temperature affect the kinetic energy of gas particles?

According to the kinetic theory of gases, temperature is directly proportional to the average kinetic energy of gas particles. As temperature increases, the particles move faster and their kinetic energy increases. Likewise, as temperature decreases, the particles move slower and their kinetic energy decreases.

What is the relationship between pressure and volume in a gas?

According to Boyle's Law, the pressure of a gas is inversely proportional to its volume at a constant temperature. This means that as the volume of a gas decreases, the pressure increases and vice versa.

How does the number of gas particles affect the pressure of a gas?

According to Avogadro's Law, at a constant temperature and volume, the pressure of a gas is directly proportional to the number of gas particles. This means that as the number of particles increases, the pressure increases and vice versa.

What is the difference between an ideal gas and a real gas?

An ideal gas is a theoretical concept that follows the assumptions of the kinetic theory of gases, such as having no intermolecular forces and occupying no volume. Real gases, on the other hand, do have intermolecular forces and occupy some volume, making them deviate from the behavior predicted by the kinetic theory of gases. However, at low pressures and high temperatures, real gases behave similarly to ideal gases.

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