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?

Answer 1

Correct option is D. $$3$$ atm
Moles of $$O_2=\dfrac{16}{32}=0.5$$

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

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

$$P_{Total}=4$$ atm

After loss of $$8$$ gm $$O_2$$

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

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

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

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

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

$$P_2=4\times 0.75$$

$$P_2=3$$ atm

Hence, option $$D$$ is correct.