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?

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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.

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Answer 1 · 2016-11-06T05:10:00

$P_{Total} = P_{Gas} + P_{SVP} = 742 mm Hg$ $P_"Total"=P_"Gas"+P_"SVP"="742 mm Hg"$

Explanation:

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

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

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

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

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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

Explanation:

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