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Gas collected over water & Fick's Law 
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#1
Jan1114, 07:45 AM

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ok, something's been bugging me
consider this simple problem: 193 mL or O_{2} was collected over water with pressure = 762 mmHg at 23°C. How many grams of oxygen were collected? The strategy is simple enough: P_{T} = P_{o2} + P_{water} P_{oxygen} = 741 mmHg →n=[itex]\frac{pV}{RT}[/itex]=7.69x10^{4} mol = 2.46 x 10^{2} g (see example in full) but every time I look at that answer, I ask myself, what about the O_{2} dissolved in the water?
the textbooks tell me to ignore it with gasses that do not dissolve appreciably in water, but then many such books (like the one in the link) ask you to calculate the O_{2} evolved... suggesting O_{2} does not dissolve appreciably in water.... but, Fick's Law and Henry's Law both tell me a significant amount of the gas WILL dissolve in the water, and if I'm not counting it, then I'm going to be off by that amount, right? I mean, we rely on Fick's law to calculate oxygen diffusion across the alveoli of the lungs! so why don't the Dalton's "gas over water" examples include the amount of oxygen in the water? (another example of this). I almost thought I had this when I assumed the P_{O2(gas)} = P_{O2(liquid)} but this would suggest I add a third term to the above equation, right/wrong? won't ignoring the dissolved gas in the water produce an underestimate? (remember the book accepts O2 as being minimally soluble but Fick and Henry both say it is)... I'd be happy to do the math, but I don't know where to start to prove this to myself, as the above equation seems incomplete, or I'm missing something. thanks so much for considering my dilemma! 


#2
Jan1114, 08:00 AM

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P: 23,395

1. There is another source of error that you ignore so far  water vapor pressure.
2. Why don't you assume some water volume and try to calculate how much oxygen can be dissolved, to be able to compare amount produced with amount "lost" to dissolving? 3. Please remember water was most likely already saturated with the oxygen before you started collecting it. How does it change the situation? 


#3
Jan1114, 08:14 PM

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#4
Jan1214, 05:21 AM

P: 16

Gas collected over water & Fick's Law
Borek, Chestermiller
Thank you for your reply, I appreciate that! but I think the Law I forgot to apply was Henry's Law (P_{gas}=k_{H}·[gas]) For O_{2} this value is 769.23 [itex]\frac{l·atm}{mol}[/itex]. if we assume P_{O2}=762 mmHg (0.975 atm) the problem becomes: [O_{2}] = [itex]\frac{0.975 atm}{769.23 \frac{l·atm}{mol}}[/itex] = 1.2675 x 10^{3} [itex]\frac{mol}{l}[/itex] if now we assume (it's not in the problem, but I add it in for accuracy) we have 1 liter of water: 1.2675 x 10^{3} mol O_{2} · [itex]\frac{32g}{1 mol}[/itex]O_{2} = 4.056 x 10^{2} g O_{2} we can now add this to our mass of oxygen that displaced the water (2.46 x ^{102} g O_{2}) but NOTICE:the dissolved oxygen is APPRECIABLY DISSOLVED IN WATER, as a matter of fact (unless I'm wrong somewhere here, and I could be), with 1 liter, we have 1.6 times as much oxygen in the water as we do in the tube! so use less water... ok, we can use 1/3rd of a liter but that still leaves 50% of the oxygen in the water ... an appreciable amount. so as I see it:
either way, it seems either the books are all doing it wrong, or I am... I'm pretty sure I am, but I can't see where I'm wrong 


#5
Jan1214, 06:56 AM

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P: 23,395

Your calculations are OK, although you still have not accounted for the fact water was already saturated with the atmospheric oxygen.
However, Purdue chemists are wrong, mass of the collected oxygen is not 2.4×10^{2} g, but 2.4×10^{1} g. I will try to contact them about it. Simple sanity check  we are not far from STP, so molar volume is not far from 24 L. 0.193 L/24 L times 32 g/mol is 0.around 26 g. We are surely slightly wrong, but not an order of magnitude, so 0.0246 is out of the question. 


#6
Jan1414, 04:13 AM

P: 16

first of all, thank you so much Borek,
I recalculated and realized there WAS an error... I can't believe I missed it the first time around! it seems the dissolution of oxygen in water turns out to be 16% of the amount of the oxygen in the graduated cylinder... still a significant amount if you were experimenting, but not as huge as it had been before. [itex]\frac{(0.975)(0.193)(32)}{(0.08206)(298)}[/itex]= 0.2462 g O_{2} in gas 0.24624 + 0.04056 = .2868 g 16.4% greaterYou're right that I have not accounted for oxygen already in the water prior to the experiment, and this should reduce that value further... If I am reasoning correctly, I can ignore ALL oxygen dissolved in water if the pressure of the surroundings matched the pressure of the system and tested everything at equilibrium... if so, this would validate the books = ignore oxygen dissolved in water as long as P_{O2_system} = P_{O2surroundings} I don't mind taking a 3 mile trip to validate the trip across the street... if that is what I just did. 


#7
Jan1414, 07:39 AM

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#8
Jan1414, 08:58 AM

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However, there is another factor to account for  how fast does the solution saturate. If the dissolution is slow enough, amount of gas dissolved can be negligible. Honestly, I have no idea how fast it is and how it should be treated. Interestingly, there is another, parallel thread that discusses very similar problem: http://www.physicsforums.com/showthread.php?t=732638 


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