How many moles of air are in the tank?

  • Thread starter mawalker
  • Start date
  • #1
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I am having trouble setting up this problem. Especially part A of the problem. I think I have a good idea of what to do for part B, but could anyone give me any hints on how to find moles from the given information?


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?
 

Answers and Replies

  • #2
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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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No, if it's a cylinder the volume is [itex]2\pi r^2 h[/itex].

[itex]\frac{4}{3}\pi r^3[/itex] is the volume of a sphere.
 
  • #5
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thanks, i always get the volume equations mixed up. using pV=nRT i got 150 atmospheres(pressure)*12315.04L(Volume)= n * .08206 (gas constant) * 303 (temperature K)

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
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I'd just convert atmospheres into pascals 1atm = 1.01x10^5 Pa, and use the standard form of the gas constant.
 
  • #8
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hmmm... that made it even worse... i took (15150000 pa)(12315.04 cm3) = n (8.3145) (303) solving for n gave me 74057610.2 ... this still seems like a lot of moles
 
  • #9
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Your volume should be in m^3.
 
  • #10
121
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Just noticed your volume is out by a factor of two anyway, since the volume of a cylinder is [tex]\pi r^2 h = \frac{\pi d^2 h}{4}[/tex]
 
  • #11
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ahhh, i'm still getting a ton of moles..... converting from cm^3 to m^3 i got 61.58 m^3....

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
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1 cm^3 = (1/100)^3 m^3 = 1E-06 m^3
 
  • #13
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ahhh, thank you....that was the problem... ok now i'm having a little bit of trouble with part B. it asks how much volume it would occupy at stp

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
1
Isn't standard pressure 1bar, i.e. 1x10^5 Pa not that it makes a difference.
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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