Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: Grey Area on 14/01/2010 20:22:46
-
OK...bear with me...
It's an old primary school experiment - you put a candle in a bowl of water, invert a gas jar over the top of it, and watch the water level rise as the oxygen is "used up" by the flame. There's a rise of about 20%, and we tell the kids that it shows that the air is 20% oxygen.
However...
A couple of years later we are going to give them Avogadro's Law, which says one mole of gas occupies a constant volume at the same temperature and pressure. Since combustion in oxygen produces one mole of Carbon Dioxide for every mole of Oxygen used...why do we get a reduction in volume?
I already know that the actual rise in practice is often only 15% or so initially - mainly because you've heated the air by burning the candle in it. I'm pretty sure if you let it cool down you'd get closer to the actual 20%. However, that still doesn't answer the question of why there is ANY drop in volume when Avogadro's Law states there shouldn't be ANY.
I also know that Avogadro's Law is an approximation as there is no such thing as an "ideal" gas...but is the variation in real life actually so high as to make a difference of 20% of such a small volume as the gas inside a gas jar?
Suggestions?
[MOD EDIT - PLEASE CAN YOU EDIT THE TITLE OF YOUR POST TO TURN IT INTO A QUESTION, WHICH IS THE POLICY ON THIS FORUM. THANKS. CHRIS.]
-
Possible explanation for the reduction in gas volume: the CO2 produced dissolves in the water.
Solubility in water of [CO2] 1.45 g/L at 25 °C, 100 kPa
http://en.wikipedia.org/wiki/Carbon_dioxide
CO2 is much more soluble in water than O2 or N2.
http://en.wikipedia.org/wiki/Carbonic_acid
-
could be me being daft but..... you warm up the gas inside the jar then as the flame goes out the gas cools causing a difference in pressure compressing the gas inside the jar sucking the water up?????
-
I think the first question is the assumption that there is a 1:1 molar ratio for O2 and CO2. waxes are basically CnH2n+2 with C being 20-40. So as products are CO2 and H2O ; the molar ratio for O2 and CO2 is more like 3:2 as is the molar ratio for O2 and H2O. Of course complete combustion needs excess O2, so we would also expect a reasonable portion of the hydrocarbon to undergo incomplete combustion and solid C would result as a product, further reducing the amount of CO2 produced. I.E. less gas in jar - and pressure equalized by the water coming in.
Doesn't fully answer,the % though. I'll have to look at it on the bench.
-
Excellent answers; thanks.
Chris
-
ok so by the burning of o2 inside the jar, there is less gass meening a drop of pressure sucking the water up. i havent quite got my head round the mole but im getting there..... i hope