Calculating Kilograms of Air in a 100m^3 Room | Molecular Weight of Air = 28g

[Bradracer18]

Here's the question:

How many kilograms of air are contained in a box 2.0m by 5.0m by 10.0m(at 1.0atm and 77 deg F)? Assume the molecular weight of air to be 28g.


This is what I've done so far, and can't come up with the correct answer.

77 deg F = 298 K
V=100m^3=100000 L
PV=nRT

(100000 L)/(298K * .0821 L*atm/mol*K) = 4087 moles

4087 moles / (28g/mole) = 146g or .146Kg air

What did I do wrong??

Thanks for the help fellas!

Brad

 

[quantumdude]

I don't see any errors. I think it might be time to call the answer you were given into question.

 

[Bradracer18]

Ok, thanks Tom. I couldn't find one either...he says you can write in your own answer if the multiple choice doesn't happen to show the correct answer...so I think that is what I shall do!

Thanks again,
Brad


***Edit...I found the problem...I took my 4087moles / 28g/mol...should have been 4087 x 28g/mol...to cancel units...and...what do you know...came up with the answer! lol...thanks again!

 

[quantumdude]

Hoo boy, I can't believe I missed that. It's time for me to get some :zzz: :zzz: :zzz:

 

[Bradracer18]

lol...not a problem...I missed it too...lol...get some sleep!

 

[jimclark]

In converting the number of mol to grams, he needs to multiply (mol x g/mol) to get grams.

The value of R (I have) is 8.31 J/(mol K). I am also using an atmospheric
pressure of 101350 Pa.

n = p V/(R T) = 101350 (N/m^2) * 100 (m^3)
----------------------------
8.31 (N m/(mol K)) * 298 K

I get, n = 4093 mol.

4093 mol * 28 g/mol = 115 kg air in 100 m^3 room.
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