1. ## Gases #2

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.

2. Originally Posted by jsu03

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.
$\displaystyle x_{O_2} = \frac{n_{O_2}}{\Sigma(n)}$

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

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

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

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

We can then put this into the ideal gas equation:

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

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