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Raoult's Law

At STP, a con...

Question

At STP, a container has $1$ mole of $A r$ , $2$ 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 $O_{2}$

Is changed by about

26 %

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Is halved

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Changes by

33 %

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Is decreased by about

8 %

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Solution

The correct option is A Is changed by about $26 %$
In first case : $n_{A r} = 1, n_{C O_{2}} = 2, n_{O_{2} = 3}$
$n_{N_{2}} = 4$
$n_{A r} + n_{C O_{2}} + n_{O_{2}} + n_{N_{2}} = 10$
$n_{O_{2}} = 3 / 10$
$∴ P_{O_{2}} = 3 P / 10$
(where $P$ is the total pressure)
In second case : $n_{O_{2}} = 2 / 9; P_{O_{2}}^{'} = 2 P / 9$
Decrease in partial pressure of $O_{2} = \frac{3 P}{10} - \frac{2 P}{9} = \frac{7 P}{90}$
% Decrease in partial pressure of $O_{2}$
$= \frac{7 P / 90}{3 P / 10} \times 100 = \frac{700}{27} = 26 %$

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At STP, a container has 1 mole of Ar, 2 moles of CO2, 3 moles of O2 and 4 moles of N2. Without changing the total pressure if one mole of O2 is removed, the partial pressure of O2

At STP, a container has $1$ mole of $A r$ , $2$ 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 $O_{2}$