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Partial Pressure Help-Please Check

  1. Aug 25, 2012 #1
    Partial Pressure Help---Please Check

    The Partial Pressure of water vapor at T = 273 K is 6.11 mbar. Find the corresponding mass concentration of water vapor. Assume P = 1 atm.


    So....

    ρH20 = (PH20*MH20/(R*T)
    ρH20= (6.11*10^2 Pa)*(18.01 g/mol)/(8.314*273 K)
    ρH20 = 11004.11/2269.722 = 4.85 g/mol

    Correct?
     
  2. jcsd
  3. Aug 26, 2012 #2

    Redbelly98

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    Re: Partial Pressure Help---Please Check

    Essentially correct, yes -- except for the units. You can either (1) think about what the units associated with "mass concentration" should be, or (2) include the units in R=8.314___?, and work out the units in the calculation. I recommend trying both approaches, in the order given here.
     
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