Question

Please Help! :)
A 3.9 L balloon filled with He at 286 K takes 4.42 days for all of the gas inside to effuse out of the balloon. Approximately how long would it take a similar balloon filled with 3.9 L of Ne under the same conditions?

Answer 1

Use D1/D2=(M2/M1)^1/2
Where D represents the rate of diffusion which may be expressed in units of volume per time

Answer 2

@aaronq

Answer 3

@cooltowl @Abmon98 gave you the formula you need, why ask someone else for help?

diffusion rate is inversely proportional to the square root of the molar mass.

Answer 4

what do the M's represent?

Answer 5

the L?

Answer 6

the molar mass
He = 4.002602 g/mol
Ne = 20.1791 g/mol

Answer 7

so would i look like this?
3.9L (x) = (4.0/20)^1/2

Answer 8

Not quite.

\[D_1=\frac{3.9\text{ L}}{4.42\text{ day}}\]\[D_2 = \text{unknown}\]
\[M_1 = 4.002602 \text{ g/mol}\]\[M_2 = 20.1791\text{ g/mol}\]

\[\frac{D_1}{D_2} = \sqrt{\frac{M_2}{M_1}}\]
When you have a value for \(D_2\), the total time for the gas to effuse out is going to be \[t = D_2*V\]where \(V = 3.9\text{ L}\)

Answer 9

Okay thank you so much for taking the time to write out every step for me:) @whpalmer4
and thank you for the equation @Abmon98

Answer 10

What do you get for your answer?

Answer 11

I got 1.53?

Answer 12

Mmm...can you show me? Remember, the Ne is a heavier molecule than the He, so it will effuse more slowly...it took 4.42 days for the He to get through, so your answer should be more than 4.42 days.

Answer 13

As the effusion rate is proportional to the square root of the molar mass, and you have multiplied the molar mass by more than a factor of 4, I would expect the effusion to take more than a factor of the square root of 4 longer, square root of 4 is 2, so answer should be more than 2*4.42 days.

Answer 14

sorry, inversely proportional to the square root of the molar mass!

Answer 15

I divided 3.9L by 4.42days for D1 getting .88. Then I divided 20.18g/mol by 4.003g/mol getting 5.04. Then I square rooted the 5.04 getting 2.25. Then I divided .88 by 2.25 to get D2 which is .3920. FInally I multiplied D2 by 3.9.

Answer 16

\[\frac{D_1}{D_2} = \sqrt{\frac{M_2}{M_1}}\]Solve that for \(D_2\)
\[D_1 = D_2 \sqrt{\frac{M_2}{M_1}}\]\[D_2 = \frac{D_1}{\sqrt{\frac{M_2}{M_1}}} = \frac{0.88}{\sqrt{\frac{20.1791}{4.002602}}} = 0.391925\]

So far, we agree. But that is a rate of liters per day effused...so we multiply that by time to get volume effused.

Answer 17

And we know the volume, but need to find the time, so

\[V = D_2 * t\]\[t = \frac{V}{D_2} = \frac{3.9}{0.391925} = \]

Answer 18

this is an example of why it is usually better to keep the units attached!

If you saw that \(D_2 = 0.391925 \text{ L/day}\) I think you would be less likely to multiply it by a number of liters and expect to get time, right?

\[0.392\frac{\text{L}}{\text{day}} * 3.9\text{ L} = 1.52851 \frac{\text{L}^2}{\text{day}}\]What kind of a unit is that, you'd be saying...

Answer 19

Whereas with the correct formula,
\[t = \frac{V}{D_2} = \frac{3.9\text{ L}}{0.391925 \frac{\text{L}}{\text{day}}} = 9.95 \text{ days}\]makes sense, right?

Answer 20

That makes sense! I tend to always leave out the measurements because it confuses but it makes perfect sense for this equation. THank you!!!

Answer 21

I always keep them when doing problems on paper, but it is a bit of a nuisance to nicely format them here, so sometimes I get lazy and drop them.

Answer 22

I would say that if it confuses you to have the units present, the right solution is to practice doing problems with the units present until they do not so. Gives good insurance against many common mistakes when there are unit conversions, brain farts, etc.

Answer 23

I will do that:) Thanks!