- #1

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A 14.0 cm-diameter compressed-air tank is 40.0 cm tall. The pressure at 30.0 C is 150 atm.

A.

How many moles of air are in the tank?

B.

What volume would this air occupy at STP?

- Thread starter mawalker
- Start date

- #1

- 53

- 0

A 14.0 cm-diameter compressed-air tank is 40.0 cm tall. The pressure at 30.0 C is 150 atm.

A.

How many moles of air are in the tank?

B.

What volume would this air occupy at STP?

- #2

- 227

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Do you know the ideal gas equation? That should be all you need for both parts.

- #3

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yes, pV = nRT.... would the volume be 4/3 pi(7)^3*(40) = 57470.2 ?

- #4

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[itex]\frac{4}{3}\pi r^3[/itex] is the volume of a sphere.

- #5

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which gave me 1847256 = 24.86n

solving for n i got n = 74293.86.... this seems like a lot of moles....

- #6

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do i need to do an additional conversion to get volume into L?

- #7

- 121

- 1

- #8

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- #9

- 121

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Your volume should be in m^3.

- #10

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- #11

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so 15150000 pa (61.58 m^3) = 93286420

(8.3145) constant * 303 degrees K = 2519.29n

divide both sides by 2519.29

and i find that n = 370288

- #12

- 121

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1 cm^3 = (1/100)^3 m^3 = 1E-06 m^3

- #13

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at stp i know:

T= 0 C or 273 K

P=1 atm or 1.01 *10^5 pa

so using pV=nRt

(1.01*10^5)V = (37.1 mol)(273 K)(8.3145 constant)

solving for V i get 3.08 m3 and this answer is not correct. i don't see what i am doing wrong here.

- #14

- 121

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Rearrange the equation before you put the numbers is in it makes it much easier. V = nRT / P, I assume you're making some silly mistake because the numbers look correct.

- #15

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thanks, it was a silly error...

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