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Dalton's Law and Partial Pressure

  1. Dec 5, 2005 #1
    1) Determine the total pressure (in atm) inside a 83.00L cylinder containing 5.00 mole of nitrogen, 6.00 mole of oxygen, and 5.00 mole of argon at 417.00 kelvin.

    P_total = (n1 + n2 + n3)*(RT/V)
    P_total = (5.00 + 5.00 + 6.00 mol)*(0.08206*417.00 K/83.00 L) = 6.60E0 atm?? (correct sig. digits?)

    2) Determine the partial pressure of nitrogen in a 6.00 L cylinder that contains 0.600 moles of oxygen, 0.600 moles of nitrogen, 0.800 moles of hydrogen, and 1.000 moles of helium at 340.00 kelvin.

    P_N = X_N*(P_total)

    X_N = (0.600 mol N)/(0.600 N + 0.600 mol O + 0.800 mol H) = 0.300

    P_total = (0.600 + 0.600 + 0.800 mol)*(RT/V)

    P_total = (2.000 mol)*(0.08206*340.00K/6.00 L) = 2.000 mol*(4.6501) = 9.30 atm

    P_N = 9.30 atm(0.300) = 2.79E0 atm ??? (correct sig. digits?)

    Thank you.
     
    Last edited: Dec 5, 2005
  2. jcsd
  3. Dec 6, 2005 #2
    Are my calculations correct?

    Thank you for any replies.
     
  4. Dec 6, 2005 #3
    Yes, they are both correct
     
  5. Dec 6, 2005 #4
    For #2, I just realized that I forgot to include the 1.000 mol He!


    Now, is it this?

    P_N = X_N*(P_total)

    X_N = (0.600 mol N)/(0.600 N + 0.600 mol O + 0.800 mol H + 1.000 mol He) = 0.200

    P_total = (0.600 + 0.600 + 0.800 mol + 1.000 mol)*(RT/V)

    P_total = (3.000 mol)*(0.08206*340.00K/6.00 L) = 3.000 mol*(4.6501) = 13.95020 atm

    P_N = 13.95020 atm(0.200) = 2.79E0 atm ??? (correct sig. digits?)

    Yet somehow I get the same answer!:redface:
     
  6. Dec 6, 2005 #5
    (I didn't check your work- I only checked your answers. :-p )
     
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