A flask contains 2.00 moles of nitrogen and 2.00 moles of helium. How many grams of argon must be pumped into the flask in order to make the partial pressure of argon twice that of helium?

Md Ariful Islam

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You can use the following law for solving this question: PfVf = P1V1 + P2V2 + P3V3 hellip; etc Pf is final pressure of the mixture Vf is Final volume of the mixture P1 = Initial pressure of gas No. 1 before mixing V1 = Initial volume of gas No. 1 before mixing P2 = Initial pressure of gas No. 2 before mixing V2 = Initial volume of gas No. 2 before mixing The same with P3 and V3 P is proportional to amount at the beginning P1V1 = P2V2 (2.00 moles of nitrogen and 2.00 moles of helium) Adding some extra argon will cause decreasing of nitrogen and helium pressure. If you have 2.00 moles of nitrogen and 2.00 moles of helium and P is proportional to amount, the amount of argon that must be pumped into the flask in order to make the partial pressure of argon twice that of helium is twice amount of nitrogen. It is 4.00 mol n = m/Mw; m = n*Mw = 4.00*40 = 160 g
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