Question

At STP, a container has 1 mole of $Ar$, 2 moles of $C{O}_2$, 3 moles of $O_2$ and 4 moles of $N_2$. Without changing the total pressure if one mole of $O_2$ is removed, the partial pressure of $O_2$

• A. is changed by about 26%
• B. is halved
• C. changes by 33%
• D. is decreased by about 8%

Answer 1

The correct option is A is changed by about 26%

In first case : $n_{Ar}=1,n_{C{O}_2}=2,n_{O_2=3}$
$n_{N_2}=4$
$n_{Ar}+n_{C{O}_2}+n_{O_2}+n_{N_2}=10$
$n_{O_2}=3/10$
$\therefore P_{O_2}=3P/10$
(where $P$ is the total pressure)
In second case : $n_{O_2}=2/9;P_{O_2}^{\prime }=2P/9$
Decrease in partial pressure of $O_2=\frac{3P}{10}-\frac{2P}{9}=\frac{7P}{90}$
% Decrease in partial pressure of $O_2$
$=\frac{7P/90}{3P/10}\times 100=\frac{700}{27}=26\%$