Partial pressure of He in the mixture of He, Ar and Ne is 0.128 atm in a total volume of 3 litres.
Explanation:
given that:
Pressure of He = 0.5 atm, volume = 1.5 L, Temperature = 298 K
Pressure of Ar = 2 atm, volume = 0.5 liter, temperature = 298K
Pressure of Ne = 1 atm, volume = 1 L, Temperature = 298 K
R = 0.0821 L atm /Kmole
Final volume of the container when all the gases are mixed = 3 litres.
temp remains constant.
Total pressure on the container=?
Frpm the formula,
P = [tex]\frac{1}{V}[/tex]
= [tex]\frac{1}{3}[/tex]
= 0.33 atm is the total pressure on the container.
The number of moles will be calculated by using PV=nRT
He
n = [tex]\frac{0.0821 x 298}{1.5 x 0.5}[/tex]
= 32 moles
Ar
n = [tex]\frac{0.0821x 298}{0.5 x 2}[/tex]
= 24.4
Ne
n= [tex]\frac{0.08206 x 298}{1x1}[/tex]
= 24.4
total number of moles = 80.8 moles
mole fraction of helium in the gas mixture = [tex]\frac{32}{80.8}[/tex]
mole fraction of He = 0.39 moles
Applying the relation:
partial pressure of He = total pressure x mole fraction
= 0.33 x 0.39
= 0.128 atm.
Partial pressure of He in the mixture is 0.128 atm.
