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!

Finding the pressure required when given number of molecules.

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    To what pressure must a piece of equipment be evacuated in order that there be only 10^8 molecules per ml at 25 degrees celcius


    2. Relevant equations
    Ideal gas equation = PV = nRT
    Charles's Law = (V2/T2 = V1/T1)

    Avagadro's number (6.02 * 10^23)

    I know that at STP one mole of gas is 22.4L
    3. The attempt at a solution

    I don't know where to begin but what I'm trying to do is find how many moles are in 10^8 molecules. Then use the Ideal Gas Equation to figure out the volume at STP and then using Charles's Law, figure out the pressure at 25 degrees or 298K.
     
  2. jcsd
  3. Nov 5, 2011 #2
    ok first you convert the number of molecules to moles using avogadros number.
    if you know that 1 mole = 6.02 * 10^23 molecules
    you need to find 'x' moles for 10^8 molecules
    Once you find the number of moles (which should be a tiny number) convert mL to L (have volume)
    You have Temperature (convert to Kelvin)
    With this you can find Pressure using the ideal gas equation assuming that R (the constant) = 0.082 L atm K−1 mol−1
    Good luck
     
  4. Nov 5, 2011 #3
    when I divide 10^8 by 6.02*10^23 I get and outrageously large number, and when I go vice versa, I still get a massive number.
     
  5. Nov 5, 2011 #4
    I get a extremely small number when i divide 10^8 by 6.02 *10^23 .... are you doing this correctly? check your steps ..
     
  6. Nov 5, 2011 #5
    oh wait, I forgot to put brackets in my calculator :P

    but these are my steps. ( I did this at STP)

    PV = nRT

    101300V = (1.66*10^-16)(8.314)(273)
    101300V =

    101300V/101300 = 3.767*10-13/101300

    v =

    P1/V1 = P2/V2

    101300/3.179*10^-18 = P2/1*10^-3 (BECAUSE ITS 1ML)

    the this leads to 2.73*10^16 which is nowhere near the right answer.

    the real answer is 4.1*10^-7pa, but I don't know how to get it :cry:
     
  7. Nov 6, 2011 #6
    you dont have to do it at STP because you are given a specific temperature, you only use STP when you are told to do so... in this case...
    T= 25 celsius (convert to Kelvin)
    and your pressure is what they are asking you... so P is your unknown....
    your volume would be 0.001 L (1ml) ; read carefully the problem 'what pressure [...] in order that there be only 10^8 molecules per ml ..that means 10^8 molecules (or 1.66*10^-16 moles) in only 1 ml (which equals 0.001 L) .. so having all these (plus constant R = 0.082 L atm K−1 mol−1 ) you can solve for P using ideal gas equation...

    post your results good luck
     
  8. Nov 6, 2011 #7
    Hmmm, I'm getting close.

    P(0.001) = (1.66*10^-16) (8.314) (298)
    P = 4.11*10^-10

    This is close but the answer is ^-7. Am I missing something here?
     
  9. Nov 6, 2011 #8
    You got it! :D your answer is in kPa( kilo pascal) 1 kPa = 10^3 Pa, so if you multiply your answer by 10^3 you get what you are supposed to... this comes from the fact that you chose R to be 8.314 and its units are L kPa K−1 mol−1 (note the kPa) ... good job! good luck
     
  10. Nov 6, 2011 #9
    but... 8.314 is Pa (according to my text book)
     
  11. Nov 6, 2011 #10
    wait what? can you post the units for constant?
     
  12. Nov 6, 2011 #11
    oh wait... the base unit for volume was m^3 so that means that volume isn't 0.001 its 0.000001 which leads to the right answer...

    I can't thank you enough, you're awesome dude, live long and prosper \m/
     
  13. Nov 6, 2011 #12
    Dude you are welcome! Good luck mate ....\m/
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the pressure required when given number of molecules.
Loading...