My textbook, as well as practically every online source I have looked into, says that when you collect gas over water, the pressure of the trapped gas plus the vapor pressure of water at that given temperature is equal to the outside atmospheric pressure, or $P_g + P_v = P_a$.

However, this doesn't make much sense to me, as the weight of the column of water below the gas is being ignored (or at least that's how it seems to me). Shouldn't the external/atmospheric pressure equal the total pressure of the trapped gas(es) above the water PLUS the pressure contributed by the weight of the water below the trapped gas (just like how the force exerted by the weight of the water column is considered in a barometer or manometer)?

Thanks in advance for helping me clear this confusion up.

  • 1
    $\begingroup$ Are the water levels inside and outside the collection vessel equal? $\endgroup$ – Chet Miller Mar 28 '19 at 0:44
  • $\begingroup$ Based on what I have seen online, the water levels are not the same (could just be a bad drawing). Does this make a difference either way? $\endgroup$ – Michael Mar 28 '19 at 1:02
  • $\begingroup$ If the water levels are the same inside and outside, the gas pressures above the inside and outside surfaces must be the same. Otherwise the gas pressure inside will not match the outside pressure. The inside gas pressure, of course, can be calculated from the hydrostatics. $\endgroup$ – Chet Miller Mar 28 '19 at 3:19
  • $\begingroup$ It is correct. Without gas there would be vacuum, or better just Pv.. When the niche P equals atm P the liquid column does not move. $\endgroup$ – Alchimista Mar 28 '19 at 11:49
  • $\begingroup$ @Michael — The level zero (for the water) is when the inner and outer pressures are equal, and so water levels evens out. If you write the exact equations for such a system, you will find out that water levels (inside and outside) are actually cancelling out, so in the end only the inner vs outter pressures matter $\endgroup$ – SteffX Mar 28 '19 at 13:34

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