(Dry) Volume Ideal Gas Law Calculation

[sandmanvgc]

Homework Statement

If the question was asking for (dry) volume, how would you do that?

Relevant Equations

In photos below

If the question was asking for (dry) volume, how would you do that?

 

[Lnewqban]

There is a container with a mix of water vapor and oxygen.
With no vapor of water present in same container, pressure of oxygen would be 1 atm at 35° F.

 

[sandmanvgc]

With no vapor of water present, pressure of oxygen would be 1 atm.

Why wouldn’t you subtract 0.555 due to water vapw from there?

 

[Lnewqban]

Please, see:
https://en.m.wikipedia.org/wiki/Dalton's_law

 

[sandmanvgc]

Please, see:
https://en.m.wikipedia.org/wiki/Dalton's_law

I’m not sure what to take from this? I see ptot is sum all gases. For taking dry volume, why wouldn’t you subtract p due to vapor h2o?

 

[Lnewqban]

Perhaps I didn't understand your original question.
Sorry.
Let's wait for other members to post.

 

[Borek]

Why wouldn’t you subtract 0.555 due to water vapw from there?

That's what they did, no? They calculated volume of the gas collected using partial pressure of the oxygen (so the volume calculated is the wet gas volume, as water @ 0.0555 atm occupies exactly the same volume, just increasing the total pressure). Dry volume would be closer to C.

Sadly, looks like whoever wrote the calculations down made a classical beginner's mistake and rounded down all intermediate numbers, so the errors accumulate and answers given are difficult to reproduce. Dry volume should be 436 mL, wet volume should be 462 mL. That's assuming 8.00 g of Ag2O, as technically 8 g has only one significant digit.

 

[Chestermiller]

Another way of doing this is to determine the total number of moles of gas first. The mole fraction of water is 0.0555, so the number of moles of water is (0.017)(0.0555)/0.9445=0.0010. So the total number of moles of gas is 0.018. This number of moles of gas is at 1 atm. See what you get for the volume of this wet gas mixture from the ideal gas law.

 

[sandmanvgc]

That's what they did, no? They calculated volume of the gas collected using partial pressure of the oxygen (so the volume calculated is the wet gas volume, as water @ 0.0555 atm occupies exactly the same volume, just increasing the total pressure). Dry volume would be closer to C.

Sadly, looks like whoever wrote the calculations down made a classical beginner's mistake and rounded down all intermediate numbers, so the errors accumulate and answers given are difficult to reproduce. Dry volume should be 436 mL, wet volume should be 462 mL. That's assuming 8.00 g of Ag2O, as technically 8 g has only one significant digit.

Why don't you subtract the pressure due to water vapor when calculating the dry volume?

 

[Borek]

Why don't you subtract the pressure due to water vapor when calculating the dry volume?

You are asking the same question for the second time. They DID subtract. Total pressure is 1 atm, dry oxygen pressure is 0.945 atm

 

[sandmanvgc]

You are asking the same question for the second time. They DID subtract. Total pressure is 1 atm, dry oxygen pressure is 0.945 atm

In the question they solve for (Wet) Volume

My original question in the topic was if instead they were asking for the (Dry) volume, what would be the difference?

Dry volume should be 436 mL, wet volume should be 462 mL

Going by your answers, it seems when you calc'ed the dry volume, difference is that you don't subtract the pressure due to water vapor from the barometric reading given in the question. I'm confused on this concept and thought if calculating the dry volume you would subtract the pressure due to H20 Vap. Going back to original question is why do you subtract for finding Wet volume but not for the dry?

 

[mjc123]

If you measured the wet volume, you would have to subtract the volume of water vapour to get the dry volume. If, as in the question, you want to calculate the wet volume, you have to add to the dry volume the volume of water vapour (which is equivalent to subtracting the vapour pressure of water).