Question

A gas mixture contains twice as many moles of o2 as n2. addition of 0.200 mol of argon to this mixture increases the pressure from 0.800 atm to 1.10 atm. how many moles of o2 are in the mixture?

Answer 1

number of moles of oxygen = x

number of moles of nitrogen = y

x = 2y

initial pressure, p1 = 0.8 atm

final pressure, p2 = 1.10 atm

At constant volume and temperature p1 / n1 = p2 / n2

=> p1 / p2 = n1 / n2

n1 = x + y = 2y + y = 3y

n2 = 0.2 + 3y

=> p1 / p2 = 3y / (0.2 + 3y)

=> 0.8 / 1.10 = 3y / (0.2 + 3y)

=> 0.8 (0.2 + 3y) = 1.10 (3y)

0.16 + 2.4y = 3.3y

=> 3.3y - 2.4y = 0.16

=> 0.9y = 0.16

=> y = 0.16 / 0.9

=. x = 2*0.16/0.9 = 0.356

Answer: 0.356 moles O2