# Gases Problem

1. Sep 29, 2007

### simplicity12

[SOLVED] Gases Problem

Hi Everyone,
I was wondering if someone can explain the following question:
An apparatus consists of three temperature-jacketed 1.000 L bulbs connected by stopcocks. Bulb A contains a mixture of H2O (g), CO2 (g), and N2 (g) at 25'C and a total pressure of 564 mm Hg. Bulb B is empty and is held at a temperature of -70'C. Bulb C is also empty and is held at a temperature of -190'C. The stopcocks are closed, and the volume of the lines connecting the bulbs is zero. CO2 sublimes at -78'C and N2 boils at -196'C.
A. The stopcock between A and B is opened, and the system is allowed to come to equilibrium. The pressure in A and B is now 219 mm Hg. what do bulbs A and B contain?

The answer is:
Bulb A contains CO2 (g) and N2 (g)
Bulb B contains CO2 (g) and N2 (g) and H2O (s).

But I have no idea how they got this. I understand that the H2O will become a solid in Bulb B since the temperature is -70'C, but why does the H2O (g) leave bulb A completely?

C. Both stopcocks are opened and the system is again allowed to come to equilibrim. The pressure throughout the system is 33.5 mm Hg. What do bulbs A, B, and C contain?

The answer is:
Bulb A contains N2 (g)
Bulb B contains N2 (g) and H2O (s)
Bulb C contains N2 (g) and CO2 (s)

I'm pretty sure that this question is similar to the previous one, but again, I don't understand why some gases completely leave one of the bulbs.
Can someone help?

2. Sep 29, 2007

### D H

Staff Emeritus
One way to look at this is to investigate what happens if the valves are only opened for a short period of time.

In the original problem, it doesn't matter whether the bulbs are excellent or poor conductors of heat because eventually the gases in a blub will come to thermal equilibrium with the bulb. I'm going to assume the bulbs are very poor heat conductors here. Suppose I open the stopcock between A and B. No work is done, and because the bulbs are poor heat conductors, there is no heat transferred to or from the gas. In other words, I've reduced the problem to a free expansion. Half of each of the constituents of gas in bulb A transfers to bulb B without any change in temperature. Note that some water vapor remains in bulb A.

Now I close the stopcock and let the system come to equilibrium. Nothing of interest happens in bulb A, but what of bulb B? As the gas cools, the vast majority of water vapor turns to snow1. The pressure drops because of the reduction temperature and the loss of gaseous H2O. The gas in bulb B turns from a mix of H2O, CO2, and N2 to a mix CO2 and N2.

What happens if I briefly open the stopcock between A and B again? What happens after I do this many times? Finally, what is the difference between my repeated open/close/wait sequence and just leaving the valve open until the system comes to equilibrium?

Notes:
1I think you are supposed to ignore that a tiny, tiny portion of water vapor remains gaseous in bulb B even after cooling to -70oC. The vapor pressure is incredibly tiny at that temperature. Treat it as zero.

3. Sep 29, 2007

### simplicity12

so there are no calculations that need to be done?

4. Sep 29, 2007

### D H

Staff Emeritus
How could there be? Unless you omitted them, uou were not supplied with any partial pressures or the volumes of the bulbs.

5. Sep 29, 2007

### simplicity12

There is, in the question it says that:
Bulb A contains a mixture of H2O (g), CO2 (g), and N2 (g) at 25'C and a total pressure of 564 mm Hg. Bulb B is empty and is held at a temperature of -70'C. And when it comes to equilibrium, the pressure in A and B becomes 219 mm Hg.

6. Sep 29, 2007

### D H

Staff Emeritus
Ahh. Yes, you can deduce the masses of the gases from problem as stated.

First look at it qualitatively. Do you understand why there is no (or essentially no) gaseous H2O in either bulb A or B after opening the first stopcock and letting the system come to equilibrium, and no gaseous CO2 in any bulb after opening both stopcocks?

To solve the problem quantitatively, work backwards and ignore the volume occupied by the little piles of H2O in bulb B and CO2 in bulb C. By "work backwards", I mean start with the end conditions and work toward the initial conditions. Finally, double check the validity of the assumption about ignoring the little snowpiles.

7. Sep 29, 2007

### simplicity12

The thing that I don't really understand is why the H2O completely leaves bulb A when the stopcock between A and B is opened...

8. Sep 29, 2007

### D H

Staff Emeritus
Did you read my first post? You didn't ask any questions about it, so I assumed you did.

9. Sep 29, 2007

### simplicity12

I read your first post. I understand why the H2O would turn into snow when it gets into the bulb B. But how come the H2O doesn't stay in Bulb A? Because, similarly, the CO2 was only in Bulb A initially, and when the stopcock opened, some of the CO2 remained the Bulb A, while some of it went to Bulb B. So why is it that Bulb A doesn't keep any of the H2O, but it keeps some of the CO2?

10. Sep 29, 2007

### D H

Staff Emeritus
Worry about the H2O only. The CO2 vanishes for the same reason.

So, you understand that H2O turns to snow in bulb B. That means there will be essentially no water vapor in bulb B. Suppose there is some water vapor in A. What happens to that water vapor? Remember, the stockcock is open between bulbs A and B.

11. Sep 29, 2007

### simplicity12

All of the water vapour from Bulb A would go into Bulb B?

12. Sep 29, 2007

### D H

Staff Emeritus
Don't guess. Why does this happen? Look at the problem in terms of repeatedly opening and closing the stopcock. All of the water vapor does not go from bulb A to bulb B (immediately). Why does all of the water eventually all end in bulb B in the solid form?

13. Sep 29, 2007

### simplicity12

:uhh: i really don't know... is it because of diffusion? Since Bulb B had no water vapour in it initially, the water vapour from Bulb A would diffuse in Bulb B when you repeatedly open and close the stopcock?

14. Sep 29, 2007

### D H

Staff Emeritus
Suppose bulb A has gaseous H2O in it only and Bulb B is at vacuum. What happens?

Now suppose you have another bulb connected to bulb B, and but this time you Maxwell's demon lies between bulb B and this new bulb instead of a dumb old stopcock. This demon lets H2O flow freely from bulb B to the new bulb but blocks the reverse flow. What happens to the water vapor? What is the difference between Maxwell's demon and deposition (deposition is the opposite of sublimation).

15. Sep 29, 2007

### simplicity12

The water vapour would go into a new bulb because it blocks the reverse flow.

16. Sep 29, 2007

### D H

Staff Emeritus
So what is the difference between the new bulb and water deposition?

17. Sep 30, 2007

### Grapz

Do u take chem at U of T

18. Oct 1, 2007

### chemisttree

Yes, it is because of diffusion. The water diffuses out of bulb A and is trapped as ice in bulb B. The remaining CO2 will diffuse into the third bulb (when that stopcock is eventually opened) and precipitate as solid CO2. You can think of the problem in terms of repeatedly opening and closing the stopcock as has been offered already or you could think of it in terms of Fickian diffusion and trapping.

19. Oct 5, 2007

### simplicity12

I understand the question now! THANK YOU VERY MUCH!

20. Oct 7, 2007

### Grapz

I dont' understand part c

Both stopcocks are opened and the system is again allowed to come to equilibrim. The pressure throughout the system is 33.5 mm Hg. What do bulbs A, B, and C contain?

The answer is:
Bulb A contains N2 (g)
Bulb B contains N2 (g) and H2O (s)
Bulb C contains N2 (g) and CO2 (s)

Why is there no H2O in bulb c?

If it were allowed to come to equilibrium, that means diffuision occurs.

Sure sum would go to bulb b, and turn to ice. But there is bound to be sum h2o gas that goes to Bulb c and also turns to ice right?

so why is there no H2O (s) in bulb c.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook