An oxygen cylinder of volume 30 L has 18.20 | vedclass

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(C) Initial moles $n_i = 18.20 \ 	ext{mol}$.Volume $V = 30 \ 	ext{L} = 30 	imes 10^{-3} \ 	ext{m}^3$.Temperature $T = 27^{\circ} 	ext{C} = 300 \

Vedclass · Accepted Answer

$0.116$

Vedclass · Answer

(C) Initial moles $n_i = 18.20 \ 	ext{mol}$.Volume $V = 30 \ 	ext{L} = 30 	imes 10^{-3} \ 	ext{m}^3$.Temperature $T = 27^{\circ} 	ext{C} = 300 \ 	ext{K}$.Gauge pressure $P_g = 11 \ 	ext{atm}$.Absolute pressure $P_{abs} = P_g + P_{atm} = 11 + 1 = 12 \ 	ext{atm}$.Using the ideal gas law $PV = nRT$, the final number of moles $n_f$ is:$n_f = \frac{P_{abs} V}{RT} = \frac{12 	imes 1.01 	imes 10^5 	imes 30 	imes 10^{-3}}{(100/12) 	imes 300} = \frac{36360}{2500} = 14.544 \ 	ext{mol}$.Moles of oxygen withdrawn $\Delta n = n_i - n_f = 18.20 - 14.544 = 3.656 \ 	ext{mol}$.Mass of oxygen withdrawn $m = \Delta n 	imes 	ext{Molar mass} = 3.656 	imes 32 \ 	ext{g} = 116.992 \ 	ext{g} \approx 0.117 \ 	ext{kg}$.Rounding to the nearest option, the mass is $0.116 \ 	ext{kg}$.

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