Candle in a Jar

[Judge Phred]

Theres a favourite teacher's experiment that involves a lighted candle standing in a bowl of water. When you place a jar over the candle and into the water, the candle goes out and the water level rises. My teacher told me it's because the oxygen gets used up.
I disagree.
Any comments?

[araya]

hmm, I've never seen that experiment. It makes sense though - the candle uses up most of the oxygen in the air trapped in the jar, which reduces the air pressure in the jar, causing the water level to rise due to different pressures acting on the surface of the water outside the jar and inside. Do you have a more plausible explanation Judge? I would say that it's possible that the increased temperature of the air inside the jar also causes an increase in the humidity of the air, which acts to put out the candle flame, but that doesn't seem like a very big effect compared to the change in pressure.

[Murray]

A very similar technique is used in a classic bar trick. I've always understood it to work pretty much like your teacher says it does. It always made sense to me.

[Quailman]

The oxygen reacts with the paraffin, beeswax, etc. through combustion to form who-knows-what conglomeration of toxic gases. If the resulting gas takes less room than the oxygen did previously, the pressure should rise. Or the result of the combustion may be particulates (smoke) that are "dissolved/suspended"? in the remaining gas and require less volume.

Does the subsequent cooling of this toxic oxygen-depleted gas result in a further rise in the water level? I don't know much about Boyle's Law.

[Ghost Post]

my understanding:

as the candle burns, it burns oxygen, and replaces it through combustion with either carbon dioxide or carbon monoxide, depending on if it's complete or incomplete combustion. (sufficient amount of oxygen leads to complete comb.; more carbon monoxide would be produced, i'd guess, as the oxygen gets less plentiful.) assuming that this smoke/carbon (?)oxide exerts less pressure than oxygen, the water level would rise, as is exhibited.

What happens as the smoke mixture cools, I would guess, is, since temperature and pressure are related directly, the water level would actually rise further as it cooled.

I think I have this right, though I'm not certain. This is from memory and it's been a while since i've played with chemistry. any arguements?

[This message has been edited by Scotty (edited 09-24-1999).]

[Judge Phred]

Here's why I disagree with the oxygen theory.

In the open flame (and with simplified chemistry):

CH4 + 2O2 = CO2 + 2H2O
Candle burns to give carbon dioxide and water vapour.
Since mol quantities of gasses occupy equal volumes, volume is INCREASING from 2 to 3 units.

You put the jar over, which captures the hot CO2, water vapour and other gasses around. The candle then goes out, gasses cool and water vapour starts to condense. This lowers the pressure, sucking up the water.

If you do it right, you can get the water over halfway up the jar. Thats a hellofalot more than the 20% oxygen in air! Also I'd bet that there is still about 15% of oxygen left in the gasses in the jar - the candle goes out very quickly.

Sorry - I'm going on a bit and I know this is not a chemistry discussion, but teachers, please give us a wink when you are oversimplifying things for an easy explanation!

[This message has been edited by Judge Phred (edited 09-24-1999).]

[araya]

Not a chemistry discussion? We discuss anything and everything around here ;-)

Isn't CH4 methane gas? I agree that paraffins are a simple hydrocarbon molecule, so I'll assume that the reaction
CH4 + 2O2 --> CO2 + 2H2O is a reasonable estimation of the reaction that occurs in a burning candle, although it's probably quite a bit different in actuality. You claim that 2 moles of gas becomes 3 moles of gas, which would be true if it was possible to build up a significant water vapour pressure at room temperature. Saturation pressure for water at 25 degrees C is just over 3 kPa (this is the maximum pressure where water is gaseous, at 25 degrees C). Now, obviously the water is heated somewhat above room temperature, so say, even at 70 C, saturation pressure is 30 kPa (this is less than a third of atmospheric pressure). The point of all these numbers is that water vapour would have to be very hot to be present at a significant pressures, or the pressure would have to be very low in order for vapour to be present at low temperatures. Since we know water vapour might come out of the candle-burning reaction at, say 100 degrees C, but it will be cooled very quickly by the surrounding gas, and the water surface via convection, we know that the vapour pressure will be very low. This means that almost all of the water coming out of the reaction will be condensing on the walls of the jar and joining the pool of water at the base. In any case, once the candle goes out, the heat is dissipated and the water vapour will drop to room temperature, meaning it will reach saturated vapour state at 3.2 kPa.

I will go into more detail here, because I'm interested in the results:

Air is 21% O2 by volume, and 79% N2 by volume (roughly). The pressure of the O2 in the air P_O2=0.21(P_air)(R_O2)/(R_air)=19 kPa (where R is the gas constant for ideal gases). P_N2=82 kPa using the same method. Note that the partial pressures add up to atmospheric pressure as we expect P_air=19+82=101kPa
The specific volume (inverse of density) of the O2 is v_O2=(R_O2)*T/(P_O2)=4.0 m^3/kg, and since x kg of O2 is replaced by (44/64)x kg of CO2, v_CO2=5.8 m^3/kg after the reaction. P_CO2=(R_CO2)*T/(v_CO2)=9.5 kPa. As mentioned previously, P_H20=3.2 kPa. So, assuming all the oxygen is used up, the air pressure after the reaction ends and everything cools will be P_air=P_N2+P_CO2+P_H2O=94.7 kPa. This pressure change would cause a change in the level of the water, equal to h=(Patm-Pair)/[(rho_water)*g], but Pair is increasing as the level of the water rises. For an ideal gas under isothermal compression, PV=constant, and the change in V is dependant on the height of the jar, which we call z (distance from water surface to top of jar, when the trapped air is at (P, T)atm. Combining everything, we get the somewhat complex expression (where P_init = 94.7 kPa)
P_atm - P_init[z/(z-h)] = rho*g*h which, when solved for h, gives h~1 cm for z=15 cm. Not nearly halfway up the jar. The only way i can think of for the water level to reach h=0.5z would be if the nitrogen started getting involved in the reaction, and some precipitates were being produced.

Um, I guess I'm done now. The explanation that the oxygen gets used up fully explains why the candle goes out, and it mostly explains the water level rising, but the cooling and condensation of the water vapour has a big part of why the pressure in the jar slowly falls after the candle has gone out. Or so I assume, having never actually seen the experiment.

[This message has been edited by araya (edited 09-27-1999).]

[Judge Phred]

Errrrgh!
I'm a little sceptical about such a precise argument without properly considering the initial assumptions, Araya! In fact that’s the problem I had in the beginning. All the theoretical analysis in the world is of little value if reality differs ...

I think you should all try this! Lets have people burning candles all around the world.

I back up my claim that this has more to do with contracting gasses than oxygen being "used up" by noting the following: it is quite easy to vary the result of this experiment by the speed with which you lower the jar.

If you do it quickly, the candle goes out after about 3 seconds and the water comes up maybe 5 to 10 percent. If you look carefully, you will see bubbles coming out as the candle goes out. I maintain that escaped gas is the main cause of the water rise when things cool down again.

If you hold the jar above the candle until is becomes hot to the touch, and then plunge it in the candle goes out almost immediately. No bubbles this time, as the gas has already escaped the jar above the candle. As the various gasses cool, condense and dissolve, the water can reach half way up the glass.

OK, the candle wouldn't burn in the first place if it weren't for the O2, but saying that this happens because the oxygen is used up is an unsatisfactory answer.

I'd say (by just looking at the real world, with not too much analysis) that the causes for this effect, in order of importance are:

1. Contraction of gasses as they cool
2. Condensation of water vapour
3. Dissolution of soluble products of combustion, such as CO and CO2

And I'll say it again - not because the oxygen got "used up"!!!

[Ghost Post]

You could try using a stronge plastic bottle with the bottom cut off it.
Get a small candle and put it floating in the water.
Now place the open bottle over the candle so that now you should
have the level of water inside and outside the bottle equal.
Close the lid of the bottle and now when the volum of the air increases
in the bottle the level of water in the bottle will be pushed down
(you will have gotten rid of the escaping gas factar).
to exclude gasses disolving in the water put a light layer of oil over the water
as long as the bottle is stronge enough it sould hold out.

I don't have any idea how this would workout but it might put your mind at rest.

Let me know how you get on

[mathgrant]

Interesting thread. BUMP.

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JDTAY and I are morons. Just ask Chess, he'll tell you.

Here's my first THP:

http://www.greylabyrinth.com/Forums/Forum5/HTML/002951.html

And here's my second:

http://www.greylabyrinth.com/Forums/Forum5/HTML/003011.html

[CzarJ]

Uh... I agree with mathgrant... I think...

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Basket-Weaving For Donuts, Where You Weave Baskets And Get Donuts.

[HyToFry]

when did araya start posting again?

[= [= [=

[Coyote]

I hope he finds his way out of the coal mine someday. I miss his chess problems.

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Gravity is a harsh mistress