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Math Help - Gases #2

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    Gases #2

    Can someone please help me with this problem Thanks.

    What volume of air contains 11.8 of oxygen gas at 273 and 1.00 . the mole fraction of oxygen gas in air is 0.21.

    I use PV=nRT where V= nRT/P and n = 11.8 g x.21 / 32 = 0.077 then I plug that and the rest into the equation but did not get the correct answer.
    Last edited by mr fantastic; October 5th 2009 at 03:10 AM. Reason: Re-titled
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  2. #2
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    Quote Originally Posted by jsu03 View Post
    Can someone please help me with this problem Thanks.

    What volume of air contains 11.8 of oxygen gas at 273 and 1.00 . the mole fraction of oxygen gas in air is 0.21.

    I use PV=nRT where V= nRT/P and n = 11.8 g x.21 / 32 = 0.077 then I plug that and the rest into the equation but did not get the correct answer.
    x_{O_2} = \frac{n_{O_2}}{\Sigma(n)}

    Where:
    • x_{O_2} is the mole fraction of O2
    • n_{O_2} is the moles of oxygen
    • \Sigma(n) is the number of moles in total


    n_{O_2} = \frac{m}{A_r} = \frac{11.8}{32} = 0.36875

    Since we now two of the above we can solve for \Sigma(n) which is the total moles of air

    \Sigma(n) = \frac{n_{O_2}}{x_{O_2}} = \frac{0.36875}{0.21} = \frac{295}{168}

    We can then put this into the ideal gas equation:

    V = \frac{nRT}{P} = \frac{\frac{295}{168} \times 8.314 \times 273}{101325} = 0.0393\,m^3 = 39.3\,dm^3 \: \text{3sf}

    edit:

    I should point out that I change P in atm to Pa so I could use the SI value of R with which I am more familiar. It does not affect the final answer and 1atm = 101325Pa so no rounding off occured
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