# Effusion, uranium enrichment

1. Apr 6, 2015

### Incand

1. The problem statement, all variables and given/known data
Naturally occurring uranium contains 0.7% $\;^{235}U$ and 99.3% $\; ^{238}U$ . To enrich it one uses a method based on repeated effusion.
a) Consider a step in this process. In a container, which is divided in two parts by a porous plug, one part contains natural uranium in the form of uranium hexaflouride $UF_6$ (gaseous). And the second part vacuum. Calculate the concentration of $\;^{235}U$ in this part when the system through continuous filling and pumping reach equilibrium.
b) How many times must the effusion process be repeated to get 20% $\;^{235}U$

2. Relevant equations
Number of collisions per second and area (where m is the mass of an individual particle)
$v^* = \frac{1}{4}n\langle v \rangle = \frac{p}{\sqrt{2\pi mkT}}$

3. The attempt at a solution
I'm assuming the form of the uranium ($UF_6$) is irrelevant and we can simply use the 99.3% and 0.7% concentrations (of mass I assume).
Assuming natural uranium enters the chamber at the same rate that the current concentration of mass leaves we have $m_{in} = m_{out}$.
$m_{238} = \frac{M}{N_A} = \frac{238}{6.02\cdot 10^{23}}$
$m_{235} = \frac{M}{N_A} = \frac{235}{6.02\cdot 10^{23}}$
I got $m_{in} = m_{out} = (m_{238out} + m_{235out}) \propto (v^*_1 + v^*_2)$. At this point I'm stuck. I don't know anything about the pressure or temperature or how many of the collisions get through the plug. Any advice on how to continue?

Last edited: Apr 6, 2015
2. Apr 6, 2015

### Staff: Mentor

The fluorine is not relevant for chemistry here, but it still influences the mass of the molecules - their relative masses get more similar (because all fluorine atoms have the same mass).

All you need are ratios, not absolute numbers. Those unknown things cancel for the ratios.

3. Apr 6, 2015

### Incand

Thanks for responding. So $UF_6$ works the same with both isotopes and what you're saying is that I should use $M_{^{235}UF_6}$ instead of $M_{^{235}U}$?

I think I understand a bit more what I should do but now i end up with the wrong answer instead.
The mass that leaves of a certain type is proportional to the concentration and root of the molecular mass with $X$ as the current concentration $m_{out} = cX\sqrt{M}$
Equilibrium
\begin{cases}0.993 = m_{238in} = m_{238out} = cX\sqrt{M_{238}} \\
0.007 = m_{235in} = m_{235out}= cY\sqrt{M_{235}}\\
1 = c(X\sqrt{M_{238}} + Y\sqrt{M_{235}} )
\end{cases}
but we have that $Y=1-X$ (set the mass that leaves/enters to 1 unit per time)
$c=\frac{1}{X(\sqrt{M_{238}} -\sqrt{M_{235}} ) + M_{235} }$
$0.993X(\sqrt{M_{238}} -\sqrt{M_{235}}) - X\sqrt{M_{238}} = -0.993\sqrt{M_{235}} \Longleftrightarrow X = \frac{0.993\sqrt{M_{235}}}{0.007\sqrt{M_{238}}+0.993\sqrt{M_{235}}}$
I end up with the new concentration being $0.702%$ using the molecular mass including the flourine $M_{^{235}UF_6} = 19*6+235$ (the $10^-3$ cancel out anyway). and $0.703%$ when using only the molecular mass for the uranium. (it should be $0.708%$ according to the book). Is there an error in the formula I get perhaps? Trying to see where I made an error.

Last edited: Apr 6, 2015
4. Apr 6, 2015

### Staff: Mentor

Right, apart from the numbers: where do 335 and 338 come from? 19*6+235 or 19*6+238 give a different result.

I don't understand your calculations, but they look too complicated. If some fraction f of 0.007 lighter molecules and some fraction f' (what is the relation between f and f'?) of 0.993 heavier atoms cross the membrane, how many lighter and heavier molecules do you have behind it? What is the concentration then?

5. Apr 6, 2015

### Incand

Sorry, should ofc be $235$ and $238$ somehow that $2$ turned into a $3$ all the way through my second post (corrected it now). Guess I'm getting a bit confused after working this problem for most of the day. Maybe I see where I go wrong now. I used proportional to $\sqrt{M}$ instead of $\frac{1}{\sqrt{M}}$. So the relation you speak of.
$\frac{f}{f'} = \frac{0.007\cdot \frac{p}{\sqrt{2\pi m_1kT} }}{0.993\cdot \frac{p}{\sqrt{2\pi m_2kT}}} = \frac{0.007\sqrt{M_2}}{0.993 \sqrt{M_1}}$. Going to think this through a bit, see If i get anywhere.

Last edited: Apr 6, 2015
6. Apr 6, 2015

### Staff: Mentor

f and f' are independent of the concentrations.
They are fractions of material that passes the membrane. You don't know the values, but f/f' is possible to calculate.

7. Apr 6, 2015

### Incand

Well the ratio should still be $\frac{f}{f'} = \sqrt{\frac{m_2}{m_2}} = \sqrt{\frac{M_2}{M_1}}$ right? But then the number of molecules leaving still has to be proportional to the amount of molecules so I should still make use of the expression I wrote earlier?
Althought i guess that's not true if the percentage I got is indeed for mass percentage and not percentage of total number of molecules (The exercise sadly only say percentage without saying percentage of what). Sorry for not understanding, not sure if this is what you meant or how to go from here (except my earlier approach.

8. Apr 6, 2015

### Staff: Mentor

f/f': right. That allows to continue with the approach of post #4.

Mass percentage / percentage of molecules: Hmm, probably mass percentage, but the difference is small (too small to be significant).

9. Apr 6, 2015

### Incand

So what I think you're getting at is that if I remove a certain concentration of the 0.007 lighther molecules i get
$c_1 = 0.007(1-f)$ left and similary
$c_2 = 0.993(1-f')$
correct?
But for equilibrium the concentration $0.007$ and $0.993$ should be $c_1$ and $c_2$.
$c_1 = c_1(1-f)$ left and similary
$c_2 = c_2(1-f')$
We're also adding matter with the concentration $0.007$ and $0.993$ at the rate we take it away. what disapears is $0.007f + 0.993f'$ which have to be equal the mass added $S$
$c_1 = c_1(1-f) +0.007S$
$c_2 = c_2(1-f') + 0.993S$
Which can be simplified too
$c_1f = 0.007S$
divide and conquer
$\frac{c_1}{c_2} \sqrt{\frac{M_2}{M_1}} = \frac{0.007}{0.993} \Longleftrightarrow \frac{c_1}{c_2}= \frac{0.007}{0.993}\sqrt{\frac{M_1}{M_2}}$
Lets call $a = \frac{c_1}{c_2} = \frac{c_1}{1-c_1} \Longleftrightarrow c_1 = \frac{a}{1+a}$ which is wrong again, the concentration is shrinking instead :( I guess I complicate things again but I really don't see how else to go about it or why I get the wrong answer (every time).

10. Apr 7, 2015

### Staff: Mentor

You are massively overthinking this.

You have 0.007 lighter molecules and 0.993 heavier ones (per molecule of matter) - let's say 7 and 993 for convenience as we are interested in the ratio only.
A fraction f of the lighter molecules will pass the membrane, so the other side has 7f of lighter molecules, and $993f'=993f \sqrt{\frac{M_1}{M_2}} < 993f$ of heavier molecules.
Nothing goes back as we are constantly keeping the vacuum on one side.
Now calculate the ratio of lighter molecules, given those two numbers (and nothing else). It should be slightly larger than 7/1000.

11. Apr 7, 2015

### Incand

At least I finally understand that I want the concentration for the molecules arriving into the vacuum (and not the other way around that I tried to calculate all this time).
Still I'm afraid I don't understand (Sorry!).
The percentage is number of moles/molecules and not mass now?
We have $7f$ lighter molecules and $993f' = 993f \sqrt{\frac{M_1}{M_2}}$ heavier molecules. Then the total amount (of molecules) would be $7f + 993f\sqrt{\frac{M_1}{M_2}}$
so to get the concentration of lighter molecules we divide $7f$ with the total
$\frac{7f}{7f+993f\sqrt{\frac{M_1}{M_2}}} = \frac{7}{7+993\sqrt{\frac{M_1}{M_2}}}$

Is this wrong? I do seem to get the last decimal wrong according to the book but I do get the exact number of iterations needed to get 20% right with this.

12. Apr 8, 2015

### Staff: Mentor

Both will be 0.007, to see a difference we would need more significant figures.
Could be a rounding issue.
Then it is fine.

13. Apr 8, 2015

### Incand

Cheers! Thanks for helping me out so much. The exact answer isn't important but that I got the right method for calculating it.

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