# Homework Help: Mass transfer

1. Nov 3, 2015

### NYK

1. The problem statement, all variables and given/known data
A (spherical) rubbery balloon of 20 cm in diameter is filed with helium. The rubber balloon wall has a thickness of 0.05 cm and diffusivity of 0.1x10-10 cm2 /s for helium. When the balloon is left in the air at 25°C, helium leaks into the air by diffusion through the rubbery wall and, as a result, the balloon shrinks. The Henry constant for helium in the rubber is 5 mol/cm3 .atm.

(1) Derive an equation that correlates the balloon size to the time;

(2) Estimate the time required for the balloon to shrink to 10 cm in diameter.

(Note: The helium pressure in the balloon is 2 atm and is essentially constant during the shrinking process. To simplify calculation, a quasi steady state can be assumed for the problem).

2. Relevant equations

J = (DH/L)(Ca - Cb)
Q = J.A.Δt
d(v(P/RT))/dt = -SJ

3. The attempt at a solution

I am having trouble starting this problem. I think that I need to somehow incorporate the mass balance with the flux equation. Then I get confused as to how I would derive the equation to equate the the balloon size as a function of time.

Does my though process sound like I am on the right track?

Thank you in advance for any help!

2. Nov 4, 2015

### Staff: Mentor

How many moles of helium are in the balloon to start with?

What is the concentration of the helium that is dissolved in the rubber on the helium side of the wall?

Chet

3. Nov 4, 2015

### NYK

Hi Chet thank you for the tips on getting started,

I used the ideal gas law (since the problem states to assume a quasi steady state) and found the number of moles of helium in the balloon initially to be:

n = PV/RT = (2 atm*33510.32cm^3)/(82.06(cm^3*atm/mol*K)*298.15K) = 2.739 mol He

Using n, I calculated the intial concentration of helium in the balloon to be:

Co = 2.739 mol/33510.32 cm^3 = 8.17 x 10^-5 mol He/cm^3

Next to find the concentration of helium dissolved in the rubber on the helium side I am trying to use:

J = (D/L)(Co-C1)

I am having trouble with finding the concentration out side of the balloon (C1)

Would i use the ideal gas law again as:

C1 = P/RT

then if so do I assume the pressure outside of the balloon to be 1 atm?

4. Nov 4, 2015

### Staff: Mentor

This was the only part that was correct.
The concentration of helium dissolved in the rubber at the interface between the helium and balloon wall is determined by using the Helium pressure in the balloon and the Henry's law constant.

What do you think the partial pressure of the helium will be in the room air after it has seeped through the wall into the room air? Do you really think it will be 1 atm.?

Chet

5. Nov 4, 2015

### NYK

Hi Chet, I am in a class right now, but just to run a thought by you before I am able to continue working on this problem, the partial pressure outside of the balloon of the helium would be 0 atm, there isnt any boundary creating a pressure when the helium escapes through the rubber walls to the outside environment.

Does that sound logical?

Then using henrys law:

P = HX

X = P/H

6. Nov 4, 2015

### Staff: Mentor

Perfect.
Actually, Henry's law is expressed as C=HP. So give me a number.

Chet

7. Nov 4, 2015

### NYK

Co = 10 mol/cm^3?

8. Nov 4, 2015

### LDavis

I am working on the same problem. Can Henry's constant be expressed as P/C=H?

9. Nov 4, 2015

### NYK

http://www.ece.gatech.edu/research/labs/vc/theory/oxide.html [Broken]

that is where i found C = HP

But I did find the same eqn you are talking about where H = P/C in the text book

Last edited by a moderator: May 7, 2017
10. Nov 4, 2015

### LDavis

Since the volume is changing and I assume the pressure is constant inside of the balloon, the mass balance gives me d(V(Pi/RT))/dt=0-S(DH/L)P and from there I assume that I can rearrange that and put S in terms of volume or radius

11. Nov 4, 2015

### NYK

S = 4πr2?

then d(V(P1/RT))/dt = -S((DH/L)P1)

dV/dt = -S(DH/L)RT

dV = -(4πr2)(DH/L)RTdt

12. Nov 5, 2015

### Staff: Mentor

Correct.

13. Nov 5, 2015

### Staff: Mentor

Yes. What is dV in terms of r and dr?

Do you think you are supposed to take into account the fact that the balloon rubber is incompressible so that $S(t)L(t)=S(0)L(0)$, or do you think they expect you to not realize that and assume that L is constant?

14. Nov 5, 2015

### LDavis

15. Nov 5, 2015

### Staff: Mentor

Oh. He mistook the value of the diameter for the value of the radius. No big deal.

Chet

16. Nov 5, 2015

### LDavis

Okay I just thought I missed something. So the idea then is to get all of the radius terms on one side of the equation to integrate from the starting radius to the final on the left and the starting and final times on the right. Does V also need to be rewritten in terms of r so we have (4/3πr^3)/(4πr^2 dr) Where the numerator is V and the denominator is S both in terms of r?

17. Nov 5, 2015

### Staff: Mentor

I don't understand your question. Can you elaborate?

Chet

18. Nov 5, 2015

### LDavis

Sorry. So the equation is dV = -(4πr2)(DH/L)RTdt and I need all of the r terms on one side so

dV = -(4πr^2)(DH/L)RTdt

dV/(4πr^2) = -(DH/L)RTdt

from here would I rewrite V in terms of r to get

(4πr^3)/(4πr^2)dr = -(DH/L)RTdt

which would simplify to

rdr = -(DH/L)RTdt

19. Nov 5, 2015

### LDavis

Sorry to take over your post NYX and thank you so much for your assistance thus far Chestermiller it is greatly appreciated

20. Nov 5, 2015

### Staff: Mentor

Looks OK, except that you differentiated the volume incorrectly.

Chet

21. Nov 5, 2015

### LDavis

I am not sure where to go from there, was my simplification of (4πr^3)/(4πr^2)dr incorrect or does it not need to be simplified?
would r^3/r^2 dr be correct, this is where I've been stuck for a day, would it be simpler to right S in terms of V so i can just leave dV as is?

22. Nov 5, 2015

### Staff: Mentor

$dV=4πr^2dr$, not $dV=4πr^3dr$