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

Solution

## The correct option is C $$3$$ atmMoles 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 $$n_1$$ )$$P_{Total}=4$$ atmAfter 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 $$\dfrac{P_1}{n_1}=\dfrac{P_2}{n_2}$$$$\dfrac{4}{1}=\dfrac{P_1}{0.75}$$$$P_2=4\times 0.75$$$$P_2=3$$ atmHence, option $$D$$ is correct.Chemistry

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