Question

Quantities of $16$g of oxygen and $14$g of nitrogen are contained in a closed bottle. The pressure inside the bottle is $4$ atm. Now, $8$g $O_2$ gas is removed from the bottle. What will be the new pressure inside the bottle?

• A. $2$ atm
• B. $4$ atm
• C. $6$ atm
• D. $3$ atm

Answer 1

Answer: (D)

$3$ atm

Moles of $O_2=\frac{16}{32}=0.5$

Moles of $N_2=\frac{14}{28}=0.5$

Total moles$=0.5+0.5=1$ (say $n_1$ )

$P_{Total}=4$ atm

After loss of $8$ gm $O_2$

$\therefore$ Moles of $O_2=\frac{16-8}{32}=0.25$

Moles of $N_2=$ Same as first $=0.5$

Total moles$=0.25+0.5=0.75$ (say $n_2$)

By using $\frac{P_1}{n_1}=\frac{P_2}{n_2}$

$\frac{4}{1}=\frac{P_1}{0.75}$

$P_2=4\times 0.75$

$P_2=3$ atm

Hence, option $D$ is correct.