Kinetic Theory - A gas mixture effuses through a hole, find the pressure change

[physconomic]

Homework Statement

So if you have a gas that is a combination of H$_2$ and HD (hydrogen-hydrogen and hydrogen-deuterium) in the proportion of 7000:1, and it effuses through a hole at constant temp into a vacuum whereby the proportion changes to 700:1, what factor would the pressure have changed?

Relevant Equations

Daltons law

$p = n_1 K_B T$
Effusion rate: $\frac{1}{4} n \bar{v}$

So I know Dalton's law as stated above which I think is applicable in this question. Then I know the effusion rate is $\frac{1}{4} n \bar{v}$, and from this we can make a differential for the time evolution of the number density of the gas in the container which is:

$\frac{dn}{dt} = \frac{A}{\phi} = \frac{A}{4}n \bar{v}$

Therefore: $\frac{dn}{n} = \frac{A\bar{v}}{4}dt$

As $n = \frac{n_0}{v}$, then the solutions must be:

$n = n_0 e^{-\frac{At}{4}n}$ for each of H$_2$ and HD

I'm not sure if this is the right way to go about this, or how to link this to get a change in the factor of the pressure in the vessel?

Thank you

 

[mjc123]

Be careful with notation; you seem to be confusing volume and velocity with your v's, and n and N₀ do not have the same dimensions as your last equation implies.

How does the effusion rate vary with the molecular weight? Can you get an expression for n(t) for each gas and proceed from there?

 

[physconomic]

Be careful with notation; you seem to be confusing volume and velocity with your v's, and n and N₀ do not have the same dimensions as your last equation implies.

How does the effusion rate vary with the molecular weight? Can you get an expression for n(t) for each gas and proceed from there?

Hi sorry about that I've edited the original post because I accidentally capitalised a few letters.

From Graham's law, $\frac{\phi_1}{\phi_2} = \sqrt{\frac{M_2}{M_1}}$

And then $n = n_0 e^{-\frac{At}{4}}\bar{v}$ for each gas?

 

[mjc123]

And what is v bar (in terms of M)?

 

[physconomic]

And what is v bar (in terms of M)?

Wouldn't it just be $\bar{v} = \sqrt{\frac{8K_B T}{\pi M}}$ as it obeys the Maxwell-Boltzmann distribution?

 

[mjc123]

Yes. So as you want to compare the two gases, which only differ in their molecular weight, simplify it to
n = n₀e^-kt/√M where k is a constant you don't need to evaluate.

 

[physconomic]

Yes. So as you want to compare the two gases, which only differ in their molecular weight, simplify it to
n = n₀e^-kt/√M where k is a constant you don't need to evaluate.

Okay, so then do I evaluate this equation for both $H_2$ and HD each? I don't understand what I would put for $n_0$ and n for each one.

 

[Charles Link]

I don't know that the proportions that are given in the OP are valid. More $H_2$ than $HD$ will show up in what effuses through. Perhaps I am missing something, but it doesn't look right to me.
Edit: Oh, I think I see it=the proportion in the original container changes to 700:1.

 

[physconomic]

I don't know that the proportions that are given in the OP are valid. More $H_2$ than $HD$ will show up in what effuses through. Perhaps I am missing something, but it doesn't look right to me.

Isn't that what the proportions say? $700 H_2$ to 1 $HD$ atom - so there is more $H_2$?

 

[Charles Link]

Please see my edit to post 8.

 

[Charles Link]

I think post 6 has it right. To add some detail: $n_1=n_{10}e^{-kt/\sqrt{2}}$, and $n_2=n_{20}e^{-kt/\sqrt{3}}$.

There is then enough info to solve for $kt$, and to get the final pressure in terms of the initial pressure. Note that the pressure of the $HD$ can be ignored for this last pressure calculation.