The capacity of a vessel is 3 , L. It contain | vedclass

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(A) The total pressure $P$ is given by the ideal gas equation $PV = n_{total}RT$,where $n_{total} = n_{O_2} + n_{N_2} + n_{CO_2}$.$1$. Calculate the n

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$5 	imes 10^5$

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(A) The total pressure $P$ is given by the ideal gas equation $PV = n_{total}RT$,where $n_{total} = n_{O_2} + n_{N_2} + n_{CO_2}$.$1$. Calculate the number of moles for each gas:$n_{O_2} = \frac{6 \, g}{32 \, g/mol} = 0.1875 \, mol$$n_{N_2} = \frac{8 \, g}{28 \, g/mol} \approx 0.2857 \, mol$$n_{CO_2} = \frac{5 \, g}{44 \, g/mol} \approx 0.1136 \, mol$$2$. Total moles $n_{total} = 0.1875 + 0.2857 + 0.1136 \approx 0.5868 \, mol$.$3$. Convert volume to $m^3$: $V = 3 \, L = 3 	imes 10^{-3} \, m^3$.$4$. Temperature $T = 27 + 273 = 300 \, K$.$5$. Calculate pressure $P = \frac{n_{total}RT}{V} = \frac{0.5868 	imes 8.31 	imes 300}{3 	imes 10^{-3}} \approx \frac{1463.3}{0.003} \approx 4.87 	imes 10^5 \, N/m^2$.Rounding to the nearest option,the pressure is approximately $5 	imes 10^5 \, N/m^2$.

The capacity of a vessel is $3 \, L$. It contains a mixture of $6 \, g$ oxygen,$8 \, g$ nitrogen,and $5 \, g$ $CO_2$ at $27^{\circ}C$. If $R = 8.31 \, J/(mol \cdot K)$,then the pressure in the vessel in $N/m^2$ will be (approx.)

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