Answer to Question #26111 in Inorganic Chemistry for salma

Question

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?

Answer 1

You can use the following law for solving this question:
P_fV_f = P₁V₁ + P₂V₂ + P₃V₃ … etc

P_f is final pressure of the mixture
V_f 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/M_w;

m = n*M_w = 4.00*40 = 160 g