Question

A closed container of volume 0.02 m${}^3$ contains a mixture of
neon and argon gases, at a temperature of 27${}^{\circ }$C and pressure
of 1 $\times 10^{5}N{m}^{-2}$ The total mass of the mixture is 28 g. If the
molar masses of neon and argon are 20 and 40 g mol
respectively, find the masses of the individual gases in the
container assuming them to be ideal (Universal gas constant
R = 8.314 J/mol-K)..

Answer 1

Given, temperature of the mixture, T = 27${}^{\circ }$ C = 300 K
Let m be the mass of the neon gas in the mixture. Then, mass
of argon would be (28 - m)
Number of gram moles of neon, $n_1=\frac{m}{20}$
Number of gram moles of argon, $n_2=\frac{(28-m)}{40}$
From Dalton's law of partial pressures.
Total pressure of the mixture (p) = Pressure due to neon (P${}_1$)
+ Pressure due to argon (p${}_2$)
$orp=p_1+p_2=\frac{n_{1}RT}{V}+\frac{n_{2}RT}{V}=(n_1+n_2)\frac{RT}{V}$
Substituting the values
$1.0\times 10^5=(\frac{m}{20}+\frac{28-m}{40})\frac{(8.314)(300)}{0.02}$
Solving this equation, we get
m = 4.074 g and 28 - m = 23.926 g
Therefore, in the mixture, 4.074 g neon is present and the rest
i.e. 23.926 g argon is present.