1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How many moles of air are in the tank?
  1. How many states (Replies: 19)

Loading...