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 gram molecular weights of neon and argon are $20$ and $40$ , respectively, then find the mass (in $\rho$ ) of argon gas in the container. Assume gases to be ideal. (Given $R=8.314\, J \,mol ^{-1} \,K ^{-1}$ )

Answer 1

Let $m$ gram be the mass of neon. Then the mass of argon is $(28-m) g$.
Total number of moles of the mixture,
$\mu=\frac{m}{20}+\frac{28-m}{40}=\frac{28+m}{40}\,\,\,...(i)$
Now, $\mu=\frac{P V}{R T}=\frac{1 \times 10^{5} \times 0.02}{8.314 \times 300}=0.8\,\,\,...(ii)$
From (i) and (ii),
$\frac{28+m}{40}=0.8 $
$\Rightarrow 28+m=32$
$ \Rightarrow m=4 g$
or mass of argon $=(28-4) \,g =24 \,g$