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Homework Help: How many moles of air are in the tank?

  1. Nov 8, 2006 #1
    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?
     
  2. jcsd
  3. Nov 8, 2006 #2
    Do you know the ideal gas equation? That should be all you need for both parts.
     
  4. Nov 8, 2006 #3
    yes, pV = nRT.... would the volume be 4/3 pi(7)^3*(40) = 57470.2 ?
     
  5. Nov 8, 2006 #4
    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.
     
  6. Nov 8, 2006 #5
    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....
     
  7. Nov 8, 2006 #6
    do i need to do an additional conversion to get volume into L?
     
  8. Nov 8, 2006 #7
    I'd just convert atmospheres into pascals 1atm = 1.01x10^5 Pa, and use the standard form of the gas constant.
     
  9. Nov 8, 2006 #8
    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
     
  10. Nov 8, 2006 #9
    Your volume should be in m^3.
     
  11. Nov 8, 2006 #10
    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]
     
  12. Nov 8, 2006 #11
    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
     
  13. Nov 8, 2006 #12
    1 cm^3 = (1/100)^3 m^3 = 1E-06 m^3
     
  14. Nov 8, 2006 #13
    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.
     
  15. Nov 8, 2006 #14
    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.
     
  16. Nov 8, 2006 #15
    thanks, it was a silly error...
     
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