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 $\times 10^5 Nm^{-2}$ 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$_2$)
$or \, \, \, p=p_1 +p_2 =\frac{n_1RT}{V} + \frac{n_2 RT}{V} =(n_1 +n_2) \frac{RT}{V}$
Substituting the values
$ \, \, \, \, \, 1.0 \times 10^5 =\bigg(\frac{m}{20} + \frac{28-m}{40}\bigg) \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.