Question

. Gaseous effusion can be used to separate isotopes. Suppose you had a balloon filled with equal molar amounts
of hydrogen, H2, and deuterium, D2, gas.  After 10 hours you find that 20.0 % of the hydrogen, H2, had effused
out of the balloon along with some unknown amount of the deuterium, D2.   
What would be the molar percentage of D2 gas in the balloon after this 10 hour period?
(Atomic Masses:         H   1.01 grams/mol;     D  2.01 grams/mol)     

Answer 1

use grahams law and find how much of the Deuterium gas effused

Answer 2

but how do i set it up>

Answer 3

\[V _{1/}V _{2}= \sqrt{M _{2}/M _{1}}\]

v1=v2

after 10 hrs you lost 20%

so v1= 80 % = 0.8

Answer 4

oh wait no

Answer 5

i haven't used the formula i forgot that v meant velocity.. i was thinking volume

Answer 6

so 20% in 10 hour .. thats 20%/10 = 2%/hr

so v1= 0.02

Answer 7

so .02/v2 = sqrt of 1/2 which is .02/v2=1/4 so v2=.08 %?

Answer 8

\[\frac{ 0.02 }{ V _{2} }=\sqrt{\frac{ 4 }{ 2 }}\]

Answer 9

V2 =0.0141

Answer 10

where did you get the 4 from?

Answer 11

Deuterium gas .. is diatomic, just like hydrogen gas. 1 deuterium atom = 2g/mol Deuterium gas 4 g/mol

Answer 12

how are we supposed to know that?

Answer 13

know what? that theres 2 molecules in the gas?

Answer 14

yes, i still dont get why its 4/2

Answer 15

\[\frac{ V _{1} }{ V _{2} }=\sqrt{\frac{ M _{2} }{ M _{1} }} \]

M = molar mass of the gas
V = velocity at which gas diffuses

notice that the order is inverted underneath the squared root sign

M2 = deuterium gas if you set V2 to be the velocity of effusion for deuterium
M1= hydrogen gas if you set V1 to be hydrogen's

Answer 16

You could also write it like this:

\[\frac{ V _{2} }{ 0.02 }=\sqrt{\frac{ 2 }{ 4 }}\]

Answer 17

as the order is arbitrary

Answer 18

but it says h=1 and d=2 , why do we double it?

Answer 19

OOOOO I GET IT LOL

Answer 20

thank you :)

Answer 21

haha okay good, no prob dude!