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Having trouble using the ideal gas law for this problem.

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A pressure as low as 1 * 10^-8 torr can be achieved using an oil diffusion pump. How many molecules are there in 1 cm^3 of a gas at this pressure if its temperature is 371 K?


    2. Relevant equations

    PV = nRT = NkT


    3. The attempt at a solution

    I converted 1*10^-8 torr to atm and got 1.31578947 * 10^-11 atm. Then, I converted 1 cm^3 to liters and got .001L. Then, I plugged all known variables into the ideal gas law. When doing PV = nRT, I got n in molecules to be 2568782.388. This was incorrect. I then tried PV = NkT, and N was also incorrect. Where did I go wrong?
     
  2. jcsd
  3. Dec 7, 2009 #2

    Andrew Mason

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    Your units are mixed up. Atmospheres are not MKS units. Convert torr to Pascals instead (Newtons/m^2)

    AM
     
  4. Dec 7, 2009 #3
    Are liters correct then? Or should that be in m^3? I converted torr to pascals, and found the molecules to be 2.602914125*10^11, which is still incorrect.
     
  5. Dec 7, 2009 #4
    lets see what I get using torrs/atm

    volume at stp = 1ml(10^-8/760)*273/371=9.78 x 10-12 ml.

    Converting to liters and dividing by 22.4 L/mole I get 4.27 x 10^-16 moles. Multiplying by A's number: 262,837,500 which should be rounded to 2.63 x 10^8. Seems like your answer is off by 1000--maybe liter m^3 conversion?
     
  6. Dec 8, 2009 #5

    Andrew Mason

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    It is rather difficult to determine where you went wrong if you do not show us your detailed calculations.

    n=PV/RT where [itex]P = 10^{-8} Torr = 1.33 \times 10^{-6} Pa[/itex] and [itex]V = 10^{-6} m^3[/itex]

    [tex]n = \frac{1.33 \times 10^{-6} \times 10^{-6}}{8.3145 \times 371} = \frac{1.33 \times 10^{-12}}{3.085 \times 10^3} = 4.31 \times 10^{-16} moles[/tex]

    Convert the moles to molecules and that is your answer.

    AM
     
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