Suppose that Daniel has a 2.00 L bottle that contains a mixture of O2, N2, and C

Question

Suppose that Daniel has a 2.00 L bottle that contains a mixture of O2, N2, and CO2 under a total pressure of 6.00 atm. He knows that the mixture contains 0.27 mol N2 and that the partial pressure of CO2 is 0.250 atm. If the temperature is 273 K, what is the partial pressure of O2?

Suppose that Daniel has a 2.00 L bottle that contains a mixture of O2, N2, and CO2 under a total pressure of 6.00 atm. He knows that the mixture contains 0.27 mol N2 and that the partial pressure of CO2 is 0.250 atm. If the temperature is 273 K, what is the partial pressure of O2? Number atm

Explanation / Answer

According to the ideal gas equation,

PV = nRT

P = 6 atm

V = 2 L

R = 0.08206 Latm/molK

T = 273K

n = PV/RT

= (6 X 2)/(0.08206 X 273K)

= 12/22.402

= 0.535 mol

Mole fraction of N2 = 0.27/0.535 = 0.505

Pressure exerted by N2 = Mole fraction of N2 X Total pressure

= 6.0 X 0.505

= 3.028 atm

Total pressure of the gas mixture = PN2 + PO2 + PCO2

PO2 = 6.0 - 3.028 - 0.25

= 2.722 atm

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