Answer:
Total pressure of the mixture = 1 atm
Partial pressure of He = 0.863 atm
Partial pressure of Ar = 0.0173 atm
Partial pressure of Ne = 0.1197 atm
We confirm that the sum of partial pressure of each gas valu
Step-by-step explanation:
STP conditions are:
1 atm of pressure and
273K of T°
We can also say that those value reffers to 1 mol.
We determine the moles of each:
5 g . 1mol / 4g = 1.25 moles of He
1g . 1mol / 39.9 g = 0.0250 moles of Ar
3.5g . 1mol / 20.18 g = 0.173 moles of Ne
Total pressure of a mixture can be obtained from the Ideal Gases Law.
P . V = n . R . T
Total moles: 1.25 + 0.025 + 0.173 = 1.448 moles
But now, we need the volume. We know that 1 mol is contained at 22.4L at STP.
1.448 mol . 22.4L / 1 mol = 32.4 L
That's the volume for our mixture. We replace:
32.4 L . P = 1.448 mol . 0.082 . 273K
P = (1.448 mol . 0.082 . 273K) / 32.4L = 1 atm
Now we can obtained the partial pressure of each gas from mole fraction
Mol of gas / Total pressure = Partial pressure of gas / Total pressure
Partial pressure of He = (1.25 / 1.448) . 1 atm = 0.863 atm
Partial pressure of Ar = (0.025 / 1.448) . 1 atm = 0.0173 atm
We know that sum of partial pressure is 1 - (Mole fraction He + Mole fr. Ar)
Partial pressure of Ne = 1 - (0.863 + 0.0173) = 0.1197 atm
We confirm that the sum of partial pressure of each gas values the total pressure of the mixture: 0.863 + 0.0173 + 0.1197 = 1