1. Oxygen (O2) is about 20% of our air, with about 80% nitrogen and a bunch of minor components. In Colorado Springs, where I grew up, the elevation is ~ 6500 feet and atmospheric pressure is only around 12 psi (vs. 14.7 psi at sea level). Under those conditions and assuming T = 300 K, calculate the number of moles of oxygen per liter, the number of oxygen molecules per liter, and the density of oxygen in g/L.
Start out with PV = nRT, plug in the numbers and solve for n = number of mols of air. Correct for 20% O2 and go from there. Post your work if you get stuck.
Here is my work:
(0.612 atm x 0.2 L)/[(0.08206 atm L/mol K) (300 K)]= 0.00497 mol O2
Final Answer:
0.00497 mol O2
0.00497 mol O2 x (6.02 x10^23 O2 molecules/1mol)= 2.99 x 10^21 O2 molecules
Final Answer:
2.99 x 10^21 O2 molecules
d=PM/RT= (0.612 atm x 32.00 g/mol)/[(0.08206 L atm/mol k)(300K)]= 0.796 g/L
Final Answer:
0.796 g/L
1 answer
When I convert 12 psi to atmosphere I don't get your value but something closer to 0.8. Making that change makes the other answers change but the procedure looks ok. I check the calculation for the first one and it is ok (except for P). I didn't check the math for the other parts but the steps are right.
1 answer