Question

A closed vessel contains 8 gm of oxgen 7 gm of nitrogen and 1 gm of hydrogen. The total pressure is 8 atmospheres at a given temperature. If now oxygen is removed without change in temperature the pressure of the remaining gas is

• A. 3 atm
• B. 4 atm
• C. 5 atm
• D. 6 atm

Answer 1

The correct option is D 6 atm

Moles of $O_2$ gas $=\frac{8}{32}=\frac{1}{4}$

Moles of $N_2$ gas $=\frac{7}{28}=\frac{1}{4}$

Moles of $H_2$ gas $=\frac{1}{2}$

Total moles $=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1$

Total moles when $O_2$ is removed $=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$

Since $V,T$ remain constant, $P\propto n$

$⟹P_{new}=\frac{3}{4}\times 8atm=6atm$