# Chemistry

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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?

• Chemistry -

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?

• Chemistry -

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.

• Chemistry -

.0820 L?

• Chemistry -

no.
Post your work and I'll find what you are doing wrong.

• Chemistry -

2.37= 1atm*22.4L/273K

• Chemistry -

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?"

• Chemistry -

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

• Chemistry -

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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