What will be the pressure exerted by a mixtur | vedclass

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Question

(A) The total pressure is calculated using the ideal gas law $pV = nRT$,where $n = \frac{m}{M}$.For methane $(CH_4)$:$n_{CH_4} = \frac{3.2 \ g}{16 \ g

Vedclass · Accepted Answer

$8.314 	imes 10^4 \ Pa$

Vedclass · Answer

(A) The total pressure is calculated using the ideal gas law $pV = nRT$,where $n = \frac{m}{M}$.For methane $(CH_4)$:$n_{CH_4} = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$.For carbon dioxide $(CO_2)$:$n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.Total moles $n_{total} = 0.2 + 0.1 = 0.3 \ mol$.Given $V = 9 \ dm^3 = 9 	imes 10^{-3} \ m^3$ and $T = 27 + 273 = 300 \ K$.Using $R = 8.314 \ J \ K^{-1} \ mol^{-1}$:$p = \frac{n_{total}RT}{V} = \frac{0.3 \ mol 	imes 8.314 \ J \ K^{-1} \ mol^{-1} 	imes 300 \ K}{9 	imes 10^{-3} \ m^3}$.$p = \frac{748.26}{0.009} \ Pa = 8.314 	imes 10^4 \ Pa$.

What will be the pressure exerted by a mixture of $3.2 \ g$ of methane and $4.4 \ g$ of carbon dioxide contained in a $9 \ dm^3$ flask at $27^{\circ} C$?

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