At a temperature of 27,^oC,a 2.461 , L vessel | vedclass

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(D) Total moles $n_{total} = n_{N_2} + n_{He} + n_{O_2} = 0.3 + 0.5 + 6.2 = 7.0 \, mol$. Using the ideal gas equation $PV = nRT$,where $T = 27 + 273 =

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$3$

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(D) Total moles $n_{total} = n_{N_2} + n_{He} + n_{O_2} = 0.3 + 0.5 + 6.2 = 7.0 \, mol$. Using the ideal gas equation $PV = nRT$,where $T = 27 + 273 = 300 \, K$ and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$: $P_{total} = \frac{n_{total}RT}{V} = \frac{7.0 	imes 0.0821 	imes 300}{2.461} = \frac{172.41}{2.461} \approx 70 \, atm$. The partial pressure of $N_2$ is given by $P_{N_2} = \chi_{N_2} 	imes P_{total}$,where $\chi_{N_2}$ is the mole fraction of $N_2$. $\chi_{N_2} = \frac{n_{N_2}}{n_{total}} = \frac{0.3}{7.0}$. $P_{N_2} = \frac{0.3}{7.0} 	imes 70 = 3 \, atm$.

At a temperature of $27\,^oC$,a $2.461 \, L$ vessel contains $0.3 \, mol$ of $N_2$,$0.5 \, mol$ of $He$,and $6.2 \, mol$ of $O_2$ gases. What is the partial pressure of $N_2$ in $atm$?

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