Asked by hanna banana

A steel container of oxygen has a volume of 20.0 L at 22 C @35 atm. What is the volume @ STP? How many mol of oxygen are in the container?
Answered by hanna banana
i got the answer, 2.37L..and 28.90 mol.. but that doesnt seem right to me, wouldnt it be the other way around?

P1V1/T1= 2.37 L
n=PV/RT= 28.90 mol

or should they be switched?
Answered by DrBob222
I worked the problem and obtained 28.916 mol which rounds to 28.9 to three s.f. but 2.37 L isn't close.
P1V1/T1 = 2.37. That is right; however, that isn't V2.
You must equate 2.37 on the left to P2V2/T2 and plug in standard conditions. As Bob Pursley wrote,
P1V1/T1 = P2V2/T2 and you have solved only half the equation.
Answered by hanna banana
.0820 L?
Answered by DrBob222
no.
Post your work and I'll find what you are doing wrong.
Answered by hanna banana
2.37= 1atm*22.4L/273K
Answered by DrBob222
Pray tell, what is the unknown. You have no unknown. In addition, what you have written is not an equality. 22.4/273 certainly isn't = 2.37.
What is the problem asking for? I thought it wanted you to calculate the new volume at the new conditions? It asks, "What is the volume at STP?"
Answered by hanna banana
Yes. i don't know why im having such a problem with this.
35atm820.0L/295K= 2.37

and the stp are, 273K, 1 atm, and 22.4 L

Answered by DrBob222
No. STP conditions are 273 K and 1 atm p. Your problem, I think, is that you have somehow convinced yourself that the volume is 22.4 L since that is the volume of a mole of gas at STP. But you don't have a mole of gas. You have already figured moles at 28.9 or so and you have much more than 1 mole of gas. You must have much more than 22.4 if you have almost 30 moles of gas.
p1 = 35 atm
V1 = 20.0 L
T1 = 273 + 22 = 295

P2 = 1 atm
V2 = unknown, solve for this.
T2 = 273

(35*20.0/295) = (1*V2/273)
Solve for V2.
(35*20.0/295) = (1*V2/273)
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