Partial Pressure Chemistry Question

Three gases, nitrogen, argon, and krypton are confined in a 10.0 L vessel at a pressure of 4.00 atm. The vessel contains 2.00 moles of nitrogen, 5.00 moles of argon, and 4.00 moles of krypton. What is the partial pressure of argon?

Partial pressure of argon = ? atm

Now suppose that 2.00 moles of oxygen are added to the vessel without changing the volume or temperature. What is the partial pressure of argon now?

Partial pressure of argon after addition of oxygen = ? atm

Now suppose that an additional 3.00 moles of argon are added to the vessel, again without any change in the volume or the temperature. What is the partial pressure of argon now?


Partial pressure of argon after addition of 3.00 more moles of argon = ? atm

mole fraction Ar = moles Ar / total gas moles = (5.00) / (11.00) = 0.455

P Ar = (mole fraction Ar)(P total) = (0.455)(4.00 atm) = 1.82 atm

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new mole fraction Ar = moles Ar / total gas moles = 5.00 / 13.00 = 0.385

When you add 2.00 moles of O2 gas into a fixed-volume container at a fixed temperature, the pressure will increase proportional to the moles of gas added, but the P Ar will remain the same. Don't believe it???

P1/n1 = P2/n2
4.00 atm / 11.00 gas moles = P2 / 13.00 gas moles
P2 = 4.73 atm

new P Ar = (new mole fraction Ar)(new P total) = (0.385)(4.73 atm) = 1.82 atm . . .no change!

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new mole fraction Ar = 8.00 moles Ar / 16.00 total gas moles = 0.500

P1/n1 = P2/n2
4.73 atm / 13.00 moles = P2 / 16.00 moles
P2 = 5.82 atm

new P Ar = (new mole fraction Ar)(P total) = (0.500)(5.82 atm) = 2.91 atm