Question

A sample of He gas (3.0 L) at 5.6 atm and 25 °C was combined with 4.5 L of Ne gas at 3.6 atm and 25 °C at constant temperature in a 9.0 L flask. The total pressure in the flask was __________ atm. Assume the initial pressure in the flask was 0.00 atm.

In a gas mixture of He, Ne, and Ar with a total pressure of 8.40 atm, the mole fraction of Ar is __________ if the partial pressures of He and Ne are 1.50 and 2.00 atm, respectively.

Answer 1

Q1-Solution-

Given-

P(He) = 5.6 atm

V(He) = 3.0 L

P(Ne) =3.6 atm

V(Ne) = 4.5 L

V2 = 9.0 L

By using Boyles law

P1V1 = P2V2

P2 = P1V1/V2

For He gas

P2(He) = (5.6 atm*3.0 L)/9.0L = 1.86 atm in 9.0L flask

For Ne gas

P2(Ne) = (3.6 atm*4.5 L)/9.0L = 1.8 atm in 9.0L flask

Dalton's Law

Total pressure = sum of partial pressures = P(He) + P(Ne) = 1.86 atm + 1.8 atm

Answer-Total pressure = 3.66 atm

Q2-Solution –

Given

Total pressure(P) = 8.40 atm

P(He) = 1.50 atm

P(Ne) = 2.00 atm

Mixture of gas He, Ne, Ar

Total pressure(P) = P(He) + P(Ne) + P(Ar)

P(Ar) = P - P(He) - P(Ne) = 8.40 atm - 1.50 atm - 2.00 atm = 4.9 atm

Partial pressure of Ar = 4.9 atm

Mole fraction = partial pressure/Total pressure = 4.9atm/8.40atm = 0.583

Answer – Mole fraction of Ar = 0.583