Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Kinetic theory of gases
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
=== Collisions with container wall === For an ideal gas in equilibrium, the rate of collisions with the container wall and velocity distribution of particles hitting the container wall can be calculated<ref name="OCW">{{cite web |title=5.62 Physical Chemistry II |url=https://ocw.mit.edu/courses/chemistry/5-62-physical-chemistry-ii-spring-2008/lecture-notes/29_562ln08.pdf |website=MIT OpenCourseWare}}</ref> based on naive kinetic theory, and the results can be used for analyzing [[Effusion#Physics in Effusion|effusive flow rate]]s, which is useful in applications such as the [[Gaseous diffusion#Technology|gaseous diffusion]] method for [[Isotope separation#Diffusion|isotope separation]]. Assume that in the container, the number density (number per unit volume) is <math>n = N/V</math> and that the particles obey [[Maxwell-Boltzmann distribution|Maxwell's velocity distribution]]: <math display="block">f_\text{Maxwell}(v_x,v_y,v_z) \, dv_x \, dv_y \, dv_z = \left(\frac{m}{2 \pi k_\text{B} T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \, dv_x \, dv_y \, dv_z</math> Then for a small area <math>dA</math> on the container wall, a particle with speed <math>v</math> at angle <math>\theta</math> from the normal of the area <math>dA</math>, will collide with the area within time interval <math>dt</math>, if it is within the distance <math>v\,dt</math> from the area <math>dA</math>. Therefore, all the particles with speed <math>v</math> at angle <math>\theta</math> from the normal that can reach area <math>dA</math> within time interval <math>dt</math> are contained in the tilted pipe with a height of <math>v\cos (\theta) dt</math> and a volume of <math>v\cos (\theta) \,dA\,dt</math>. The total number of particles that reach area <math>dA</math> within time interval <math>dt</math> also depends on the velocity distribution; All in all, it calculates to be:<math display="block">n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_\text{B}T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right).</math> Integrating this over all appropriate velocities within the constraint <math>v > 0</math>, <math display="inline">0 < \theta < \frac{\pi}{2}</math>, <math>0 < \phi < 2\pi</math> yields the number of atomic or molecular collisions with a wall of a container per unit area per unit time: <math display="block">J_\text{collision} = \frac{\displaystyle\int_0^{\pi/2} \cos(\theta) \sin(\theta) \, d\theta}{\displaystyle\int_0^\pi \sin(\theta) \, d\theta}\times n \bar v = \frac{1}{4} n \bar v = \frac{n}{4} \sqrt{\frac{8 k_\mathrm{B} T}{\pi m}}. </math> This quantity is also known as the "impingement rate" in vacuum physics. Note that to calculate the average speed <math>\bar{v}</math> of the Maxwell's velocity distribution, one has to integrate over <math>v > 0 </math>, <math>0 < \theta < \pi </math>, <math>0 < \phi < 2\pi</math>. The momentum transfer to the container wall from particles hitting the area <math>dA</math> with speed <math>v</math> at angle <math>\theta</math> from the normal, in time interval <math>dt</math> is: <math display="block">[2mv \cos(\theta)]\times n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_\text{B}T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right).</math> Integrating this over all appropriate velocities within the constraint <math>v > 0</math>, <math display="inline">0 < \theta < \frac{\pi}{2}</math>, <math>0 < \phi < 2\pi</math> yields the [[pressure]] (consistent with [[Ideal gas law]]): <math display="block">P = \frac{\displaystyle 2\int_0^{\pi/2} \cos^2(\theta) \sin(\theta) \, d\theta}{\displaystyle \int_0^\pi \sin(\theta) \, d\theta}\times n mv_\text{rms}^2 = \frac{1}{3} n mv_\text{rms}^2 = \frac{2}{3} n\langle E_\text{kin}\rangle = n k_\mathrm{B} T </math> If this small area <math>A</math> is punched to become a small hole, the [[Effusion#Physics in Effusion|effusive flow rate]] will be: <math display="block">\Phi_\text{effusion} = J_\text{collision} A= n A \sqrt{\frac{k_\mathrm{B} T}{2 \pi m}}. </math> Combined with the [[ideal gas law]], this yields <math display="block">\Phi_\text{effusion} = \frac{P A}{\sqrt{2 \pi m k_\mathrm{B} T}}. </math> The above expression is consistent with [[Graham's law]]. To calculate the velocity distribution of particles hitting this small area, we must take into account that all the particles with <math>(v,\theta,\phi)</math> that hit the area <math>dA</math> within the time interval <math>dt</math> are contained in the tilted pipe with a height of <math>v\cos (\theta) \, dt</math> and a volume of <math>v\cos (\theta) \, dA \, dt</math>; Therefore, compared to the Maxwell distribution, the velocity distribution will have an extra factor of <math>v\cos \theta</math>: <math display="block">\begin{align} f(v,\theta,\phi) \, dv \, d\theta \, d\phi &= \lambda v\cos{\theta} \left(\frac{m}{2 \pi k T}\right)^{3/2} e^{- \frac{mv^2}{2k_\mathrm{B} T}}(v^2\sin{\theta} \, dv \, d\theta \, d\phi) \end{align}</math> with the constraint <math display="inline">v > 0</math>, <math display="inline">0 < \theta < \frac{\pi}{2}</math>, <math>0 < \phi < 2\pi</math>. The constant <math>\lambda</math> can be determined by the normalization condition <math display="inline">\int f(v,\theta,\phi) \, dv \, d\theta \, d\phi=1</math> to be <math display="inline">4/\bar{v} </math>, and overall: <math display="block">\begin{align} f(v,\theta,\phi) \, dv \, d\theta \, d\phi &= \frac{1}{2\pi} \left(\frac{m}{k_\mathrm{B} T}\right)^2e^{- \frac{mv^2}{2k_\mathrm{B} T}} (v^3\sin{\theta}\cos{\theta} \, dv \, d\theta \, d\phi) \\ \end{align};\quad v>0,\, 0<\theta<\frac \pi 2,\, 0<\phi<2\pi</math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)