It is not a fixed mass of air in the glass...
As the candle burns and the air expands, air bubbles escape through the wine.
See the video at the 42 second mark up to the 50 sec mark- vigorous bubbling!
Those bubbles are on the inside of the glass, not on the outside. You'll also notice that, at the time shown, the wine is already being pulled INTO the glass, indicating that there is a vacuum in the glass. You can also see a surface tension formed step in the wine, as it thins down right next to the glass.
Those bubbles are just air/CO2/gasses being pulled out of the wine by the vacuum.
As the candle burns and the air expands, air bubbles escape through the wine.
This is the key. As the candle burns in the glass, there are two competing effects.
1) the air is expanding because it is heating.
2) O2 is being converted into CO2, but as you point out, 38 molecules of O2 turn into 25 molecules of CO2. Those two molecules take up identical amount of space, so (in the erroneous assumption that the candle burns out all of the O2), 21% of the air (the O2 component) turns into .21(25/38) = .14 = 14% of the air (in the form of CO2). You lose (21% - 14%) = 7% of the gas.
Now, a perfect vacuum will lift water about 33'. 7% vacuum will lift water .07 x 33' x 12"/' = 28". Wow.
But, in reality, MUCH less (only about 10%) of the O2 gets converted before the flame goes out.
The flame extinguishes when the O2 concentration drops to around 19%. see:
https://www.picotech.com/library/results/burning-oxygen
So this drops the 7% down to (2/21) x .07 = .007 = 0.7% & the potential lift to just 2.8". Enough to pull the wine into the glass.
And this lift (& the 0.7% vacuum created by O2 -> CO2 conversion) is slightly offset (i.e., reduced) by the heating of the gasses, as you mentioned.
I was curious as to which effect would win the race at the start, when the glass is first placed over the candle. It seems clear from the video that no air is pushed out of the glass due to thermal expansion, and the conversion of O2 to Co2 (& the slight pressure drop that results) wins the race, right from the start.
At the instant the flame burns out there is a fixed volume of (nitrogen + CO2 + water vapor + argon) at atmospheric pressure. As the gas mix cools, it contracts to a smaller volume as wine comes into the glass.
Nope. While the flame is still burning (well before the flame goes out) there is already a vacuum in the glass due to the conversion of O2 to CO2. That's seen by the lifted wine & those vacuum-generated bubbles, both occurring inside the glass.
__
BTW, the points of my questions were:
- The glass to plate seal can NOT be water-tight, otherwise the wine couldn't flow into the glass.
- Since it isn't water-tight, it certainly can NOT be air-tight.
- Not being air-tight, it is entirely possible for heated, expanded air to get pushed out of the glass.
But it is clear from the video that this didn't happen.
And the effect is due solely to the conversion of just a small percent (~10%) of the O2 into a smaller amount of CO2.
__
So, this doesn't apply much at all to the primary cause of the pressure loss in footballs, the temp drop.
With one slight exception that we haven't discussed very much.
There is one slight effect that that this might mimic: The condensation of some of the water vapor into liquid water as the air inside the balls cooled from ~ 75°F to ~50°F. If this happened, then it is exactly like some gaseous oxygen in the glass being converted into liquid water.
And if some water vapor gets turned into liquid water, then n (in "nR/V") is, again, not quite a constant.
Thanks for this, PBPF.
This was fun.