Given the following, what is the pressure of dry hydrogen gas if the vapor pressure of water at 25°C is 3.8 mm Hg?

Hydrogen gas produced in the laboratory by the reaction of zinc and hydrochloric acid as collected over water at 25°C. The barometric pressure at the time was 742.5 mm Hg.

1 Answer
anor277
Nov 6, 2016

P_"Total"=P_"Gas"+P_"SVP"="742 mm Hg"

Explanation:

And P_"SVP" is the so called "saturated vapour pressure", which for this experiment = "3.8 mm Hg"

So P_"dihydrogen" = "742 mm Hg "-" 3.8 mm Hg = 738.2 mm Hg".

Note that here the quoted value of P_"SVP" is wrong. I suspect that they got confused with respect to a pressure measurement in "mm Hg" and "Pascal".

The vapour pressure of water at 298*K is "24 mm Hg".

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