[Kinetic Theory] Speed of diffusion.

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Homework Statement


Find the mean speed of the molecules escaping through a hole of area [itex]\sigma[/itex]. The vessel has volume [itex]V[/itex] and the molecules mass [itex]m[/itex].


Homework Equations


[itex]dN_w^e = \frac {dA}{4V} w \frac{dN_w}{dw} dw[/itex] [itex]w[/itex] is the speed.

[itex]dN_w = \frac {4N}{\sqrt \pi c^3} w^2 \exp {\frac {-w^2}{c^2} }dw[/itex], which if [itex]c=\sqrt {frac {2KT}{m}} is the Maxwell-Boltzmann Distribution Function.


The Attempt at a Solution



Well, the first thing I did was integrate the escapes molecules through the area... as I found no dependence of the speed and the area (is this correct?), getting

[itex]dN_w^e = \frac {\sigma}{4V} w \frac{dN_w}{dw} dw_e[/itex]

Then using a M-B distribution for the molecules

[itex]dN_w^e = \frac {\sigma}{4V} w \frac {4N}{\sqrt \pi c^3} w^2 \exp {\frac {-w^2}{c^2} } dw[/itex]

[itex]dN_w^e = \frac {\sigma}{V} \frac {N}{\sqrt \pi c^3} w^3 \exp {\frac {-w^2}{c^2} } dw[/itex]

The if I want to find the mean speed it should be

[itex]\bar w_e = \frac {1}{N_e}\int w dN_w^e [/itex] where [itex]N_e[/itex] should be the number of escaping molecules right? But as I wasn't figuring out how to resolve that so I divided by the total number molecules of the system. ----PLEASE WRITE ME SOME WORDS ABOUT THIS---

[itex]\bar w_e = \frac {\sigma}{\sqrt \pi c^3 V} \int_0^{\infty} w^3 \exp {\frac {-w^2}{c^2} } dw [/itex] which I think it is

[itex] \bar w_e = \frac {\sigma}{\sqrt \pi c^3 V} \frac{3\sqrt {\pi c^5}}{8} = \frac {3 \sigma c^2}{8V} = \frac {3}{8} \frac {\sigma}{V} \frac {2kT}{m} [/itex]

Which has units of acceleration, no speed. Then I realized that the distribution I was getting was molecules exiting the vessel per unit of time which made me conclude that the calculation made sense but in that case I've no idea how to estimate the mean speed of the molecules exiting the vessel.

Excuse me if the errors are quite stupid, I couldn't read much about Kinetic Theory until now.

Thank you for your time!!!
 

Answers and Replies

  • #2
TSny
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[itex]\bar w_e = \frac {1}{N_e}\int w dN_w^e [/itex] where [itex]N_e[/itex] should be the number of escaping molecules right? But as I wasn't figuring out how to resolve that so I divided by the total number molecules of the system. ----PLEASE WRITE ME SOME WORDS ABOUT THIS---
##dN_w^e## is your distribution function for number of particles escaping at speed w per unit time. As you said, you need to "normalize" by dividing by the total number of particles escaping (at all speeds) per unit of time, not the total number of particles in the system. What would you do to ##dN_w^e## to get the total number escaping per unit time?
 
  • #3
TSny
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[itex]\bar w_e = \frac {\sigma}{\sqrt \pi c^3 V} \int_0^{\infty} w^3 \exp {\frac {-w^2}{c^2} } dw [/itex] which I think it is

Do you have the correct power of ##w## in the integrand?
 
  • #4
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You are rigut, it's [itex]w^4[/itex] but the result of the integral is right anyway.

So I should divide by the integral of [itex]dN_w^e[/itex] over all speeds, which I think it is:

[itex]\frac {\sigma N}{sqrt \pi V c^3} . \frac {c^4}{2} = \frac {\sigma N c}{2 \sqrt \pi V} [/itex]

The I have

[itex] \frac { \frac {\sigma N } {\sqrt \pi V c^3} \int_0^{\infty} w^4 \exp{\frac {-w^2}{c^2} } dw} { N_e^w }[/itex]

which is

[itex] \frac { \frac {3 \sigma N c^2}{8 V} }{ \frac {\sigma N c}{2 \sqrt \pi V } } = \frac {6 \sqrt \pi c}{8} = \sqrt {\frac {18 \pi k T} {16 m} } = \sqrt {\frac {9 \pi k T}{8 m}}[/itex]


Which now, AT LEAST, has units of speed. Is this result okey? I find weird the independence of both area and volume.


Thanks a lot for the help TSny
 
Last edited:
  • #5
TSny
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Yes, that looks correct to me.
 

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