The pressure of a mixture of 64 g of oxygen, | vedclass

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(D) $1$. Calculate the number of moles $(n)$ for each gas: $n_{O_2} = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$ $n_{N_2} = \frac{28 \ g}{28 \ g/mol} = 1 \

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$\frac{2 P}{3}$

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(D) $1$. Calculate the number of moles $(n)$ for each gas: $n_{O_2} = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$ $n_{N_2} = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$ $n_{CO_2} = \frac{132 \ g}{44 \ g/mol} = 3 \ mol$ $2$. The total number of moles initially is $n_{total} = 2 + 1 + 3 = 6 \ mol$. $3$. According to the ideal gas law, $PV = nRT$. Since $V, R, T$ are constant, $P \propto n$. $4$. Therefore, $P = k 	imes 6$, where $k$ is a constant. $5$. After removing oxygen, the remaining moles are $n_{remaining} = 1 (N_2) + 3 (CO_2) = 4 \ mol$. $6$. The new pressure $P'$ is $P' = k 	imes 4$. $7$. Taking the ratio: $\frac{P'}{P} = \frac{4}{6} = \frac{2}{3}$. $8$. Thus, $P' = \frac{2 P}{3}$.

The pressure of a mixture of $64 \ g$ of oxygen, $28 \ g$ of nitrogen, and $132 \ g$ of carbon dioxide gases in a closed vessel is $P$. Under isothermal conditions, if the entire oxygen is removed from the vessel, the pressure of the mixture of the remaining two gases is:

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