Suppose that Daniel has a 2.00L bottle that contains a mixture of O2,N2, and CO2 under a total pressure of 4.50atm.

Question

He knows that the mixture contains 0.310 mol N2 and that the partial pressure ofCO2 is 0.250atm. If the temperature is 273 K, what is the partial pressure of O2?

Answer 1

Given:

Mixture of N₂, O₂ and CO₂

Volume = 2.0 L

Pt = 4.50 atm

0.310 mol N₂

P_CO2 = 0.250 atm

T = 273K

Asked for:

P_O2

The approach will be to first find the total number of moles of gas that are present. Then we will find the mole fraction of each gas. To obtain partial pressure of O₂, we will multiply mole fraction O₂ by total pressure.

PV = nRT

n = PV/RT = (4.50 atm)(2.00 L) / (0.0821 Latm/Kmol)(273K)

n = 0.402 moles of gas present

mole fraction N₂ = 0.310 mol N2 / 0.402 mol = 0.771

mole fraction CO₂ = 0.250 atm/4.50 atm = 0.0556

mole fraction O₂ = 1.000 - 0.771 - 0.0556 = 0.1734

partial pressure O₂ = 0.1734 x 4.50 atm = 0.780 atm

partial pressure N2 = 0.771 x 4.50 atm = 3.47 atm

partial pressure CO2 = 0.0556 x 4.50 atm = 0.250 atm

As a check, all these partial pressures do add up to 4.50 atm.