Question

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

Answer 1

Answer:

(a) $V_B=11.68L$

(b) $x_{He}=0.533$

Explanation:

Hello,

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

$n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol$

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

$P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar$

The moles of helium could be computed via solver as:

$n_{He}=2.358mol$

Or algebraically:

$3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol$

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

$V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\ \\V_B=11.68L$

Finally, he mole fraction of He is:

$x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533$

Regards.