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 N_{2}(g) , n = 2.00 mol

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

number of moles of O_{2}(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 N_{2}(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 O_{2}(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 N_{2} (g) is , pN_{2} =
total pressure x Mole fraction of N_{2}(g)

= 4.30 atm x 0.57

= 2.45 atm

Therefore the partial pressure of N_{2}(g) is 2.45
atm

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Please show all step by step work. Thank You!
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