If 1 , L of N_2 and 7/8 , L of O_2 gases are | vedclass

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(C) For the same temperature $T$ and pressure $P$,according to Avogadro's law,$n \propto V$. $	herefore \frac{n_{N_2}}{n_{O_2}} = \frac{V_{N_2}}{V_{O

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$m(N_2) = m(O_2)$

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(C) For the same temperature $T$ and pressure $P$,according to Avogadro's law,$n \propto V$. $	herefore \frac{n_{N_2}}{n_{O_2}} = \frac{V_{N_2}}{V_{O_2}} = \frac{1}{7/8} = \frac{8}{7}$. Since $n = \frac{m}{M}$,where $m$ is mass and $M$ is molar mass: $\frac{m(N_2) / M(N_2)}{m(O_2) / M(O_2)} = \frac{8}{7}$. Given $M(N_2) = 28 \, g/mol$ and $M(O_2) = 32 \, g/mol$: $\frac{m(N_2) / 28}{m(O_2) / 32} = \frac{8}{7}$. $\frac{m(N_2)}{m(O_2)} = \frac{8}{7} 	imes \frac{28}{32} = \frac{8}{7} 	imes \frac{7}{8} = 1$. Therefore,$m(N_2) = m(O_2)$.

If $1 \, L$ of $N_2$ and $7/8 \, L$ of $O_2$ gases are mixed at the same temperature and pressure,what is the relationship between their masses in the mixture?

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