Question

A system contains three gases: 2.00 moles of N2(g), 1.00 moles of He(g), and 0.50 moles of O2(g). What is the partial pressure of N2(g) if the total pressure of the system is 4.30 atm?

PLEASE help by showing steps not sure where I went wrong!

Answer 1

Given number of moles of N₂(g) , n = 2.00 mol

number of moles of He(g) , n' = 1.00 mol

number of moles of O₂(g) , n'' = 0.50 mol

Total number of moles , N = n + n' + n''

                                          = 2.00+1.00+0.50

                                          = 3.50 mol

Mole fraction of N₂(g) , X = n / N

                                         = 2.00 / 3.50

                                        = 0.57

Mole fraction of He(g) , X' = n' / N

                                         = 1.00 / 3.50

                                        = 0.29

Mole fraction of O₂(g) , X'' = n'' / N

                                         = 0.50 / 3.50

                                        = 0.14

According to Raoult's law

Partial pressure of each gas = total pressure x mole fraction

Partial pressure of N₂ (g) is , pN₂ = total pressure x Mole fraction of N₂(g)

                                                     = 4.30 atm x 0.57

                                                     = 2.45 atm

Therefore the partial pressure of N₂(g) is 2.45 atm