A sealed flask with a capacity of 2 , dm^{3} | vedclass

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(C) $1$. Calculate the number of moles of propane $(C_{3}H_{8})$: Molar mass of $C_{3}H_{8} = (3 	imes 12) + (8 	imes 1) = 44 \, g \, mol^{-1}$. Mol

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$1655$

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(C) $1$. Calculate the number of moles of propane $(C_{3}H_{8})$: Molar mass of $C_{3}H_{8} = (3 	imes 12) + (8 	imes 1) = 44 \, g \, mol^{-1}$. Moles $(n)$ = $\frac{11 \, g}{44 \, g \, mol^{-1}} = 0.25 \, mol$. $2$. Use the ideal gas equation $PV = nRT$: Given $P = 2 \, MPa = 2 	imes 10^{6} \, Pa$,$V = 2 \, dm^{3} = 2 	imes 10^{-3} \, m^{3}$,$n = 0.25 \, mol$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$. $2 	imes 10^{6} 	imes 2 	imes 10^{-3} = 0.25 	imes 8.3 	imes T$. $4000 = 2.075 	imes T$. $T = \frac{4000}{2.075} \approx 1927.71 \, K$. $3$. Convert temperature to Celsius: $T(^{\circ}C) = T(K) - 273.15 = 1927.71 - 273.15 = 1654.56 \, ^{\circ}C$. Rounding to the nearest integer,we get $1655 \, ^{\circ}C$.

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