I was revising my chemistry and while doing a problem on collecting a gas over water, I felt like I disagreed with a part of their explanation.

Firstly, an illustration to show how I think about pressure, I could be wrong about this but from what I researched it seems correct:


A barometer, it's pretty self explanatory. At the bottom level the atmospheric pressure and the pressure in the mercury are equalized. Everywhere at this level the pressure is the same.

In the vacuum tube this pressure causes the mercury to push against the vacuum, being able to hold a column of mercury 760mm high. At the top of the mercury in the vacuum tube the pressure is 0; if it wasn't 0, it would be able to push some more mercury higher.

Now the collecting gas over water part, I will first explain how they do it in the textbook:

collecting gas over water textbook

They explain that the gas collected in the inverted bottle, is equalized with atmospheric pressure. Then they also say that this gas collected over the water, contains some water vapor. So the pressure of the gas is actually equal to: "atmospheric pressure" - "pressure due to water vapor".

I posit that the pressure of the gas mixture in the bottle is not exactly equal to atmospheric pressure:

collecting gas over water diagram

If my diagram is correct, the pressure at the bottom of the bottle is equal to atmospheric pressure but the pressure at the water surface in the bottle is equal to "atmospheric pressure" - "pressure due to the column of water".

I calculated how much this "pressure due to the column of water" would be for a bottle 300mm high, and it would come down to around 2900 Pa. Which is quite small, about 3% of atmospheric pressure. So maybe they don't mention this small distinction because the error is quite small anyway? The lower the column of water gets the lower the error becomes.

I would appreciate someone telling me if I'm thinking about it correctly or if I'm totally wandering into the fog. Thanks.


1 Answer 1


You are correct. The atmospheric pressure equals the pressure due to the water column plus the pressure due to the collected gas and water vapour.

$$P_{gas} = P_{atm} - P_{column} - P_{w.vap} $$

You can take the surface of the water in the trough as datum and that would lead you to the above equation.

  • $\begingroup$ Strange that it's wrong in most sources then, I googled a bit for "collecting gas over water" and everything I came across did it like that. $\endgroup$ Commented Jun 21, 2022 at 17:58
  • $\begingroup$ chem.libretexts.org/Bookshelves/General_Chemistry/… Here the calculation has been performed correctly. $\endgroup$
    – Mechanic
    Commented Jun 21, 2022 at 18:20
  • $\begingroup$ It doesn't seem to be the same problem, atmospheric pressure does not support a column of water in the diagram on that page, so Pcolumn is 0. And in the example exercise they also do not account for the column. There will always be a column present except if Pgas + Pwatervapor = Patm. Maybe when you collect a gas over water they wait until the gas is at the water level. Because then indeed Pgas + Pwatervapor = Patm. In some exercises in my textbook they specifically talk about this pressure equivalency even when there is a column present. $\endgroup$ Commented Jun 21, 2022 at 19:06
  • $\begingroup$ @imjert Ulens Mechanic's answer, and your hypothesis, are both correct. Where the text books may appear to be misleading is that a water column of 1 Atmosphere has a height of 10330 mm (compared with 760 mm for Mercury). So for a collection tube of typical dimensions found in a laboratory (say 300 mm high), the column pressure is no more than ~0.03 atmospheres - and less as the column drops - so to a reasonable approximation the internal pressure of a gas collected over water equals the external pressure. $\endgroup$
    – Penguino
    Commented Jun 21, 2022 at 22:06
  • $\begingroup$ @Penguino, yeah that's what I was thinking as well. But... they do account for the pressure due to water vapour. Which also accounts for a very small amount (also around 0.03 atmospheres at 25 degrees C). So they do account for one small discrepancy, but not the other, while they both cause an error that is about equal in size. $\endgroup$ Commented Jun 22, 2022 at 12:10

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