Question

Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The

Ar in compartment A, which has a volume VA of 6.00 L, has a pressure of 2.00 bar. The He in
compartment B of unknown volume V3 has a pressure of 5.00 bar. When the two compartments
are connected and the gases allowed to mix, the total pressure of gas is 3.60 bar. Assume both
gases behave ideally
(a) [4 marks) Determine the volume of compartment B.
(b) [2 marks] Determine the mole fraction of He in the mixture of gases.

Answer 1

(a)
`V_B=11.68L`

(b)
`x_(He)=0.533`

Step-by-step explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

`n_A=(0.082(atm*L)/(mol*K)*298K)/(1.974 atm*6.00L)=2.063mol`

Thus, since the final pressure is 3.60 bar, we can write:

`P=x_(Ar)P_A+x_(He)P_B\\\\P=(n_(Ar))/(n_(Ar)+n_(He)) P_A+(n_(He))/(n_(Ar)+n_(He)) P_B\\\\3.60bar=(2.063mol)/(2.063mol+n_(He)) *2.00bar+(n_(He))/(2.063mol+n_(He)) *5.00bar`

The moles of helium could be computed via solver as:

`n_(He)=2.358mol`

Or algebraically:

`3.60bar=(1)/(2.063mol+n_(He)) *(4.0126+5.00*n_(He))\\\\7.314+3.60n_(He)=4.013+5.00*n_(He)\\\\7.314-4.013=5.00*n_(He)-3.60n_(He)\\\\n_(He)=(3.3)/(1.4)=2.358mol`

In such a way, the volume of the compartment B is:

`V_B=(n_(He)RT)/(P_B)=(2.358mol*0.082(atm*L)/(mol*K)*298.15K)/(4.935atm)\\ \\V_B=11.68L`

Finally, he mole fraction of He is:

`x_(He)=(2.358)/(2.358+2.063)\\ \\x_(He)=0.533`

Regards.