[omgwtfitsp]

Hi guys, I have a question that related about the partial pressure of hydrogen gas but I do not know how to approach it at all.

"In an experiment 125 mL of hydrogen gas is collected over water. If the atmospheric pressure is 0.983 atm and the temperature is 295 K, what is the partial pressure of hydrogen gas?"

So the only thing I know now is:

V = 0.125 L
T = 295 K or 22 degrees Celsius
P = 0.983 atm or 99.6 kPa (atmospheric pressure)
p.p = ?

The only thing I know is that when a question asks about partial pressure, it usually means there is another gas involved that is also exerting some pressure. So according to John Dalton's Law, total pressure equals the sum of the pressure of all the gases. But what is the other gas? Or maybe I don't even use this? I'm too lost, and my teacher never taught this to us.

How do I go about solving this? If anyone can help that would be greatly appreciated, because it would help me on my upcoming test.

[Borek]

What gas is ALWAYS in equilibrium with liquid water?

Note you will need to use tables to solve this question, it doesn't contain all the necessary information.

[omgwtfitsp]

What gas is ALWAYS in equilibrium with liquid water?

Note you will need to use tables to solve this question, it doesn't contain all the necessary information.

I just started to learn this so I'm not so sure what you are talking about. Are you talking about the vapour pressure of water table?

So in the question it tells you the temperature is 22 degrees Celsius so when I look up the chart it gives 19.8 mm Hg ( or about 2.64kPa ) of vapour pressure. Would I just take the atmospheric pressure that was given (0.983 atm/99.6 kPa) and subtract the vapour pressure? Does that leave the partial pressure of hydrogen?

So like: The atmospheric pressure - pressure of vapour = partial pressure of hydrogen?
                     99.6 Kpa - 2.64 Kpa = 96.96 kPa

So is the partial pressure of hydrogen 96.96 kPa?

[Borek]

Wasn't that hard

[omgwtfitsp]

Wasn't that hard

Oh sweet! Thanks so much.