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Thread: Ideal gas equation, fins volume

  1. #1
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    Ideal gas equation, fins volume

    Two tanks (A and B) are connected together by a small pipe which contains a valve.
    Initially the valve is closed and tank A, which has a volume of 1 m3 contains air
    at a pressure of 6 atm and a temperature of 60C. Tank B contains a mixture of
    nitrogen and oxygen (95 mol% N2) at 12 atm pressure and 90C .
    The valve is then opened and the contents of the two tanks mix. After complete
    mixing has taken place, the mixed gas contains 85 mol% N2. Calculate the volume
    of tank B.

    Consider that air contains 79 mol% nitrogen and 21 mol% oxygen and that the
    volume of any gas at 273 K and 1 atm is 22.414 m3 kmol��1 - alternatively, 1 atm
    = 101.325 kPa, R = 8.3143 kJ kmol��1 K��1.



    I know I have to make use of the ideal gas equation, however not sure how to start, what is my unknown?
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  2. #2
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    Quote Originally Posted by Tweety View Post
    I know I have to make use of the ideal gas equation, however not sure how to start, what is my unknown?
    I think I have worked it out, but I am not sure if this is the correct method/answer, so if anyone could check my working, would greatly appreciate.

    pv = nRT

    Tank A:

    $\displaystyle v = 1 m^{3} $

    $\displaystyle P = 6 atm $

    $\displaystyle n=? $

    $\displaystyle N_{2} $ : $\displaystyle 0.79 mol\% $

    $\displaystyle O_{2} $ $\displaystyle , 0.21mol\% $

    $\displaystyle T= 333k $

    Tank B:

    $\displaystyle n=?$

    $\displaystyle N_{2} $ : $\displaystyle 0.95 mol\% $


    $\displaystyle O_{2} $ : $\displaystyle 0.05mol\% 0_{2} $

    $\displaystyle p= 12 atm $

    $\displaystyle T = 363k $

    $\displaystyle V=? $

    Tank C:

    $\displaystyle N_{2} : 0.85 mol\% $

    $\displaystyle O_{2} : 0.15mol\% $


    1 atm = 101325 pa
    6 atm = 607950 pa

    $\displaystyle moles \\\\\ in\\\\\\\\\\ tank A : 607950 \times 1 = n \times 8.3143 \times 333 $

    n = 219.6 moles

    $\displaystyle N_{2}: (219.6 \times 0.79 ) + 0.95B = 0.85c $

    $\displaystyle O_{2} : (219.6 \times 0.21) + 0.05B = 0.15 C $

    solve the simultaneous equation for 'B' moles, and than use that value in the Pv= nRT to work out the volume, which I got to be 0.327.
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