Question

A closed container of volume $30litre$ contains a mixture of nitrogen and oxygen gases, at a tempereture of $27$ and pressure of $4atm.$ The total mass of the mixture is $148gm.$ The moles of individual gases in the container are (Take $R=0.08litreatm/moleK$)

• A. $n_{N_2}=2moles,n_{O_2}=3mole$
• B. $n_{N_2}=3mole,n_{O_2}=2mole$
• C. $n_{N_2}=4mole,n_{O_2}=1mole$
• D. $n_{N_2}=2.5mole,n_{O_2}=2.5mole$

Answer 1

Answer: (B)

$n_{N_2}=3mole,n_{O_2}=2mole$

$p=4atm$, $v=30litres$, $T=300K$, $R=0.08litreatm/moleK$

By ideal gas equation,

$pv=n_{T}RT$

$\therefore n_T=\frac{pv}{RT}=\frac{4\times 30}{0.08\times 300}=5$

$\therefore n_{N_2}+n_{O_2}=5$

Let, $x$ be the weight of $N_2$ in grams

$\therefore$ weight of $O_2=(148-x)$grams

$\therefore \frac{x}{28}+\frac{(148-x)}{32}=5$

$\therefore 32x+4144-28x=4480$

$\therefore 4x=336$

$\therefore x=84g$

$\therefore$ No. of moles of $N_2\left(n_{{NO}_2}\right)=\frac{84}{28}=3$moles

& No. of moles of $O_2\left(n_{O_2}\right)=\frac{148-84}{32}=2$moles