Kinetic Theory Speed distribution (general form)

In summary, the equation for finding the average of a function, g, that depends on the speed and angle of a molecule is: <g(v,θ)> = (1/2)<v^2> * <g(v,θ)> / <v>
  • #1
Robsta
88
0

Homework Statement


A molecule has a velocity v and speed v. I've worked out (and understand) that the number of molecules in a gas with speeds between v and v+dv and moving at angles between Ө and Ө+dӨ to any chosen axis is: (1/2)nf(v)dvsin(Ө)dӨ The internet verifies this. f(v) is the speed distribution (often Maxwellian but this question is general).

Here's the question. If u is a cartesian component of v (eg v_x), show by integration that:
<u> = 0,
<u^2>=(1/3)<v^2>
<mod(u)> = (1/2)<mod(v)>

Homework Equations


The question hints that <v> is the integral from 0 to infinity of vf(v)dv divided by the same integral of f(v)dv and think I understand why. But I'm not sure how to put it in terms of u?

The Attempt at a Solution


Intuitively <u> = 0 because there are the same number of molecules going "left" as there are going "right". But I can't show it rigorously using the equation above. I've found a couple of cases where f(v) is the maxwellian distribution but this question specifically asks you to solve it generally. I keep getting myself in a muddle with the infinite limits as well.

The question is attached as a picture. I've taken photos as well of some jottings. Thanks in advance to anybody even just for some hints.
 

Attachments

  • 20150105_213157.jpg
    20150105_213157.jpg
    38.2 KB · Views: 396
  • 20150105_220000.jpg
    20150105_220000.jpg
    38.6 KB · Views: 385
  • 20150105_220042.jpg
    20150105_220042.jpg
    35 KB · Views: 378
Physics news on Phys.org
  • #2
Hello, and welcome to PF!

Suppose that you have some property, g, of a molecule that depends on the speed v and angle θ that the velocity of the molecule makes to the z axis. So, g = g(v,θ). How would you find the average of g(v,θ) over all the molecules of the gas? That is, how would you find <g(v,θ)>? What integral (or integrals) would you need to perform?
 
Last edited:
  • #3
I suppose I'd have to integrate g with respect to every possible velocity dv and angle dθ?
Attached to this is a solution that I've tried to each part of the question, but I've fudged part i and parts ii and iii are wrong and I how now idea why :(
 

Attachments

  • 20150106_142111.jpg
    20150106_142111.jpg
    40.9 KB · Views: 403
  • #4
I've spoken to a lot of people about this and they've all tried to help but none of them really seem to understand the question. Specifically what exactly is mod(v) when v is already the modulus of a vector??
 
  • #5
You have the right idea. When doing <vz>, how would you express vz as a function of v and θ? The θ part of vz will contribute to the integrand for the integration over θ.
 
  • #6
Yep, v_z = v*sin (theta) and the sin goes into the sin integral and the v goes into the v integral. For part a you integrate sin squared and for part b you integrate sin cubed. I've found this so far to give the wrong answer though. I think it's all in the file I uploaded above, can you see it?
 
  • #7
Robsta said:
Yep, v_z = v*sin (theta) ...

Oops, check that.
 
  • #8
Incredible. You're right, v_z = v cos(theta). I've just solved all three parts. I'm so grateful to you for doing this, I'm going back to uni after some time off and this is part of my first hand in. I really want to make a good impression and because of you I can. I'll make sure I spend some time on this site and pay it forward. Attached is the full answer for anybody that might stumble across this in the future.

Thanks again.
 

Attachments

  • 20150106_150630.jpg
    20150106_150630.jpg
    42.9 KB · Views: 400
  • #9
Good work!
 

What is the Kinetic Theory Speed distribution?

The Kinetic Theory Speed distribution is a mathematical model that describes the distribution of speeds of particles in a gas. It is based on the principles of kinetic theory, which states that the average kinetic energy of gas particles is directly proportional to their temperature.

What is the general form of the Kinetic Theory Speed distribution?

The general form of the Kinetic Theory Speed distribution is given by the Maxwell-Boltzmann distribution, which is a probability distribution function that describes the speed of particles in a gas. It is given by the equation f(v) = 4π(μ/2πkT)^3/2 * v^2 * e^(-μv^2/2kT), where μ is the molar mass of the gas, k is the Boltzmann constant, T is the temperature, and v is the speed of the particle.

What factors affect the Kinetic Theory Speed distribution?

The Kinetic Theory Speed distribution is affected by three main factors: temperature, molar mass, and the number of particles in the gas. An increase in temperature will result in a higher average speed of particles, while an increase in molar mass will result in a lower average speed. Additionally, the number of particles in the gas affects the overall shape of the distribution.

What is the significance of the Kinetic Theory Speed distribution?

The Kinetic Theory Speed distribution is important because it helps us understand and predict the behavior of gases at the molecular level. It allows us to calculate important parameters such as the average speed, root-mean-square speed, and most probable speed of gas particles. This information is crucial in many scientific fields, including thermodynamics, atmospheric science, and chemical kinetics.

How does the Kinetic Theory Speed distribution compare to real gases?

The Kinetic Theory Speed distribution is a theoretical model and does not perfectly describe the behavior of real gases. In real gases, there are deviations from the predicted distribution due to intermolecular forces and particle collisions. However, the Kinetic Theory Speed distribution provides a good approximation for most gases under normal conditions and is a useful tool for understanding gas behavior.

Similar threads

  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
3K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top