Gases #2

jsu03 · #1 $Old$ October 4th, 2009, 12:37 PM

Can someone please help me with this problem Thanks.

What volume of air contains 11.8

of oxygen gas at 273 ${\rm K}$ and 1.00 ${\rm atm}$ . 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.

e^(i*pi) · #2 $Old$ October 4th, 2009, 12:50 PM

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 ${\rm K}$ and 1.00 ${\rm atm}$ . 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