Question

If 6.3 moles of O2 and 8.15 moles of N2 are placed in a 65.1 L tank at a
temperature of 348 C, what will the pressure of O2?

Answer 1

Answer:

Partial pressure of O2 = 4.93 atm

Explanation:

PV = n RT        R = .082057 L-atm/mol-K

n = 6.3 + 8.15 = 14.45 moles

348 C = 621.15 K

P (65.1) = 14.45 ( .082057)(621.15) = 11.31 atm

Now the fraction of pressure from Oxygen is

   6.3 / (6.3+8.15)  * 11.31 = 4.93 atm