Question

A Compartment A holds helium at pressure 5 atm, and compartment B holds argon at pressure 10atm. the partition between the gases is removed and the gases are allowed to mix. The mole fraction of helium in the mixture is 0.40 and the total volume of both compartments is 30.0 L . Temperature of helium and argon before and after mixing was 500K. Find volume of compartment A and B.

Answer 1

Given:

Pressure of He = 5 atm

Pressure of Ar = 10 atm

Mole fraction of He = 0.40

Total volume = 30.0 L

Temperature of before and after mixing = 500 K

Solution:

Calculation of mole fraction of Ar

Mole fraction of Ar = 1 – Mol fraction of He = 1 – 0.40

= 0.60

Partial pressure of He = mol fraction x Total pressure

Total pressure = partial pressure / mol fraction

= 5 atm / 0.40

= 12.5 atm

Calculation of total moles

n = PV / RT

= 12.5 atm x 30.0 L / ( 0.08206 L atm per K per mol x 500 K )

= 9.13 mol

Now Number of moles of He

= 9.13 mol x 0.40 = 3.65 moles

Volume of He gas (A) compartment = n RT / p

= 3.65 mol x 0.08206 L atm per K per mol x 500 K / 5 atm

= 30 L

Volume of B compartment

Partial pressure of Ar = Total pressure x mole fraction

= 12.5 atm x 0.60

= 7.5 atm

Number of moles of Ar = 9.13 mol x 0.60 =5.49 mol

Volume of B compartment

= 5.49 mol x0.08206 L atm per K per mol x 500 K / 7.5 atm

=30.0 L

Volume of B compartment = 30.0 L