Question

At STP, a container has $3$ moles of $He,3$ moles of $O_2$ and $4$ moles $N_2$. Without changing total pressure, if $2$ moles of $O_2$ is removed, the partial pressure of $O_2$ will be decreased by:

• A. $26.66\%$
• B. $40.33\%$
• C. $58.33\%$
• D. $66.66\%$

Answer 1

The correct option is C $58.33\%$

Let total pressure of mixture is $P_T$.
The partial pressure of oxygen is $P_{O_2}=\frac{3}{10}\times P_T$
After removing $2$ moles of $O_2$, the partial pressure of oxygen is $P_{O_2}^{\prime }=\frac{1}{8}\times P_T$
$\therefore$ Decreasing in partial pressure of $O_2$
$=\frac{\frac{3P_T}{10}-\frac{P_T}{8}}{\frac{3P_T}{10}}\times 100=58.33\%$