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Mixing ratio Check please

  1. Aug 25, 2012 #1
    A typical global concentration of hydroxii radicals (OH) is about 10^6 moleculues/cm^3. What is the mixing ratio corresponding to this concentration at sea level P = 10^5 Pa and T=298 K?


    Using the previous equations derived earlier

    ndOH= (P*Av/R*T)*COH
    (ndOH*R*T/(P*Av) ) = COH

    (10^6 molecules/cm^3)*(1 cm/100 m)*(1 cm/100 m)*(1 cm/100 m) = 1 molecule/m^3

    ((1 molecule/m^3 )*(8.314 J/mol*K)*(298 K)/(10^5 Pa)(6.022*10^23)) = COH

    2477.572/6.022*10^23= 4.11*10^-21

    Is this process correct? Thanks.
     
  2. jcsd
  3. Aug 25, 2012 #2
    There are 100 cm per meter not the other way around.
     
  4. Aug 25, 2012 #3
    Thanks but otherwise the correct process right?
     
  5. Aug 25, 2012 #4
    [itex] PV= nRT = NkT [/itex]
    where N is number of particles, k is boltzman's constant, n is number of moles, and R is the gas constant.

    [itex] C_{particles} = N/V = P/{kT} [/itex]

    [itex] mixing ratio = C_{OH}/C_{total particles} = N_{OH}/N_{total particles} [/itex]

    [itex] N_{OH}/N_{total particles} = {N_{OH}}/{{PV}/{kT}} = {10^6}/(({10^5}*{.01^3})/({1.381*10^{-23}}*{298}}) = 4.115*10^{-14} [/itex]

    You can keep doing it in moles if you want to but it just adds an extra step when you are being given the number of particles already. Looks like you did it right except for the unit conversion.
     
    Last edited: Aug 26, 2012
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