Question

A mixture containing 0.770 mol He(g), 0.244 mol Ne(g), and 0.119 mol Ar(g) is confined in a 10.00-L vessel at 25 ∘C

Part A

Calculate the partial pressure of He in the mixture.

Part B

Calculate the partial pressure of Ne in the mixture

Part C

Calculate the partial pressure of Ar in the mixture.

Part D

Calculate the total pressure of the mixture

Answer 1

Total moles = 0.770 + 0.244 + 0.119 = 1.133 mols

Pressure of mixture P = nRT/V

with,

n = 1.133 mol

R = gas constant

T = 25 + 273 = 298 K

V = 10.0 L

we get,

P = 1.133 x 0.08205 x 298/10 = 2.77 atm

Part A : Partial pressure of He in the mixture = mole fraction of He x total pressure

mole fraction of He = 0.770/1.133 = 0.68

So partial pressure of He = 0.68 x 2.77 = 1.884 atm

Part B : Partial pressure of Ne in the mixture = mole fraction of He x total pressure

mole fraction of Ne = 0.244/1.133 = 0.21

So partial pressure of Ne = 0.21 x 2.77 = 0.582 atm

Part C : Partial pressure of Ar in the mixture = mole fraction of He x total pressure

mole fraction of Ar = 0.119/1.133 = 0.11

So partial pressure of Ar = 0.11 x 2.77 = 0.305 atm

Part D : Total pressure of mixture = 2.77 atm