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Calculating the total and partial pressure of gas mixture

  1. Aug 23, 2015 #1
    1. The problem/question is as follows:

    1 mole of O2 mixed with N2 gas (PN2= 5 atm at 10 degrees celcius in a 1 L flask. What is the total pressure after 2 moles of gas is allowed to escape? How about the partial pressure of O2?

    [tex]R= 0.08206\frac{L atm}{mol K}[/tex]



    2. Relevant equations
    Using the ideal gas law, I have calculated the quantity of moles of nitrogen gas in the mixture, I add one mol of oxygen and it results in 1.2153 moles total. The problem states that 2 moles of gas are allowed to escape, yet there is not even 2 moles of gas in the system. We can't have negative moles so the situation seems impossible, or a poorly designed question at best.


    3. The attempt at a solution

    Ptot= ∑PO2 + PN2

    nN2=[tex]\frac{PV}{RT}[/tex]

    = [tex]\frac{(5)(1)}{(0.08206)(283)}[/tex]

    = 0.2153 moles N2

    ntot = 1 + 0.2153 = 1.2153 moles total

    PN2= Ptot [tex]/frac{nN2}{ntot}[/tex]

    [tex]5= Ptot\frac{0.2153}{1.2153}[/tex]

    [tex]\frac{5}{0.1772}[/tex]= Ptot= 28.22 atm

    PO2= 23.22 atm


    These values seem to make sense, but I'm not sure how to reconcile for how 2 moles are escaping.

    Interestingly, if I subtract 2 moles from the 1.2153 and in corporate that into the procedure as ntot, it will result in a negative total pressure of -18 atm, and also interestingly result in PO2= -23.22 atm (the opposite of the result above)


    Any information regarding how 2 moles can escape out when there is only 1.2153 moles contained?

    PS Sorry about the coding, it for some reason isn't working very well.
     
  2. jcsd
  3. Aug 23, 2015 #2
    Or I'm thinking the answer could simply be zero for both cases because all particles of gas have escaped, all the moles are gone and there's no pressure in the flask in that case.
     
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