A container holds 8 , g of O_2 and 7 , g of N | vedclass

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(D) The total pressure $P$ in the container is the sum of the partial pressures of $O_2$ and $N_2$ according to Dalton's Law: $P = P_{O_2} + P_{N_2}$.

Vedclass · Accepted Answer

$5$

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(D) The total pressure $P$ in the container is the sum of the partial pressures of $O_2$ and $N_2$ according to Dalton's Law: $P = P_{O_2} + P_{N_2}$.Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$:$P = \left( \frac{m_{O_2}}{M_{O_2}} + \frac{m_{N_2}}{M_{N_2}} ight) \frac{RT}{V}$.Given $m_{O_2} = 8 \, g$,$M_{O_2} = 32 \, g/mol$,$m_{N_2} = 7 \, g$,$M_{N_2} = 28 \, g/mol$,and $P = 10 \, atm$:$10 = \left( \frac{8}{32} + \frac{7}{28} ight) \frac{RT}{V} = \left( \frac{1}{4} + \frac{1}{4} ight) \frac{RT}{V} = \frac{1}{2} \frac{RT}{V}$.Thus,$\frac{RT}{V} = 20 \, atm$.When $O_2$ is removed,the remaining pressure is due only to $N_2$:$P' = \frac{m_{N_2}}{M_{N_2}} \frac{RT}{V} = \frac{7}{28} 	imes 20 = \frac{1}{4} 	imes 20 = 5 \, atm$.

$A$ container holds $8 \, g$ of $O_2$ and $7 \, g$ of $N_2$ at a pressure of $10 \, atm$. If $O_2$ is removed,the pressure becomes ... $atm$.

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