[Frederick95]

I have to use the formula PV=nRT and I am given the partial pressure of water vapour (1.94 kPa) and the atmospheric pressue (102.6 kPa)

I have collected oxygen gas by the downward displacement of water.This oxygen was produced in the first 60 s of the reaction.

What must I do to find Pressure?

I rearranged P = nRT but that does not work since I am not given mols.

[Borek]

You did it again - you have started new thread with the same question. Be warned.

I believe you were already answered - total pressure is sum of partial pressures, that is

P = 102.6 = PO₂ + PH₂O = PO₂ + 1.94

If you still don't know how to solve for PO₂... I give up.

[Arkcon]

Frederick95:
This question is the same one as this thread here:
http://www.chemicalforums.com/index.php?topic=24068.msg91212#msg91212
And probably half a dozen of the same topic that you've started. 

I'm not counting the posts of Marcus Soutlo: http://www.chemicalforums.com/index.php?action=profile;u=19672;sa=showPosts
who may be in the same class as you, or may be your sock puppet (that's for the MODs to decide), at any rate you can browse his questions as well, and try to gain info from those posts as well. 

There's no need to start thread after thread, most of us browse the latest posts thread, so we don't miss any post.  The MOD's job is to read every post for abuse, so no one's missed you.  Anyone can click on your name, and get your latest posts: http://www.chemicalforums.com/index.php?action=profile;u=19571;sa=showPosts
and see where your thought process is coming from.

Now Borek: has directly told you, what I was hoping you'd draw out yourself, that the vapor pressure of water you've collected the gas over is adding pressure to the gas and must be compensated for in PV=nRT, to use the observed V to calculate the correct n.  This is pretty pertinent these days, there's at least one, and perhaps two other posts that require measuring the gas collected over water.  I'm hoping those posters are reading these articles of yours, and you could read theirs, and see if becomes clearer.

Now, the original post of yours referenced above, I ask:

Partial pressure of water vapour:1.94 kPa.

Now, what's this part for?

And you respond with statements like:

The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2)

Since this is a pressure of the substance in the gas phase which is established at a particular temperature,

The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is oxygen (O2), at 290 K?

The partial pressure of water vapour (1.94 kPa) represents the pressure of the gas phase of water , which is H20.

Sorry, oxygen cannot be water since it has no hydrogen to perform hydrogen bonding

These statements are, in fact, wrong.  And they're not little mistakes of understanding -- you've copied random text, like "hydrogen bonding" or threw in a temperature, when you called the vapor phase of water, oxygen gas.  That's not work, as much as it took you to write, that's random text copying from your textbook to keep a thread alive. 

I actually called you a 'bot' -- an automated text reading posting program, such things are common in the net.  You're probably not a 'bot', but you do act like one, you're only slightly more sophisticated than an auto-email program.

Next time people stop posting answers to your questions, stop and read all the posts, including your own question -- I did have to tell you twice, 1 ml=1cm³, in the original post.  You may gain benefit from printing out the text, and working on it and the problem at a desk, away from the hum of a CPU fan and the lights of a monitor, to see what you have to do to solve a problem.  And to ask well thought out questions about what you're missing.

[Marcus Soutlo]

AHA!

So....

P = 102.6
P = PO2 + PH2O
P = PO2 + 1.94
Reararranging gives:

PO2 = P(atmospheric pressure) - 1.94


?

[Borek]

OK