# Gas contained in bottle

1. May 19, 2006

### Amith2006

Sir,
An open mouthed bottle contains a gas at 60 degree Celsius. To what temperature should the bottle be heated so that ¼th of the mass of the gas may leave?
I think that by increasing the temperature the gas molecules gain enough energy to overcome the intermolecular forces of attraction. But I don’t know how to relate them mathematically. Please help.

2. May 19, 2006

### Andrew Mason

PV=nRT

AM

3. May 19, 2006

### Amith2006

The volume and the number of moles of the gas are not given. How to calculate the temperature?

4. May 19, 2006

### Andrew Mason

PV=nRT

There are two ways to approach this. Consider it as equivalent to a volume expansion to 4/3 of the original volume, 1/4 of which is outside the bottle, or consider it as a loss of 1/4 of the number of molecules from a fixed volume.

P is constant. R is constant. In the latter case, V is constant and nf is 3/4 of ni. So how must T change if PV = nRT still holds? In the former case, n is constant but Vf = 4Vi/3. Again, work out how T must change in order for the ideal gas law to hold?

AM

5. May 19, 2006

### Gokul43201

Staff Emeritus
I think AM meant to say 5/4, not 4/3.

Edit: In light of subsequent posts, please disregard the above line.

Last edited: May 20, 2006
6. May 19, 2006

### Andrew Mason

Actually, I meant 4/3. I originally thought 5/4 but changed it. The volume expanding to 4/3 original volume leaves 3/4 of the original gas in the bottle.

AM

7. May 20, 2006

### Gokul43201

Staff Emeritus
Yes, that's right...I take that back. The answer is clearly 4/3. As for an approach to solve the problem, I find making V -> 4V/3 less intuitive than making n -> 3n/4 (at fixed V, and ignoring the escaped gas). Nevertheless, by the first approach, my error was in forgetting about the thermal expansion of the escaped gas (by a factor of 4/3).