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(A) The absolute pressure $P$ is given by $P = P_{gauge} + P_{atm}$. Assuming $P_{atm} = 1 \; atm = 1.013 	imes 10^{5} \; Pa$.Initial absolute pressu

Vedclass · Accepted Answer

$0.13$

Vedclass · Answer

(A) The absolute pressure $P$ is given by $P = P_{gauge} + P_{atm}$. Assuming $P_{atm} = 1 \; atm = 1.013 	imes 10^{5} \; Pa$.Initial absolute pressure $P_{1} = 15 + 1 = 16 \; atm = 16 	imes 1.013 	imes 10^{5} \; Pa = 1.6208 	imes 10^{6} \; Pa$.Initial temperature $T_{1} = 27 + 273 = 300 \; K$.Volume $V = 30 \; L = 30 	imes 10^{-3} \; m^{3}$.Using the ideal gas equation $PV = nRT = (m/M)RT$,the initial mass $m_{1} = (P_{1} V M) / (R T_{1})$.$m_{1} = (1.6208 	imes 10^{6} 	imes 30 	imes 10^{-3} 	imes 32 	imes 10^{-3}) / (8.31 	imes 300) \approx 0.624 \; kg$.Final absolute pressure $P_{2} = 11 + 1 = 12 \; atm = 12 	imes 1.013 	imes 10^{5} \; Pa = 1.2156 	imes 10^{6} \; Pa$.Final temperature $T_{2} = 17 + 273 = 290 \; K$.Final mass $m_{2} = (P_{2} V M) / (R T_{2})$.$m_{2} = (1.2156 	imes 10^{6} 	imes 30 	imes 10^{-3} 	imes 32 	imes 10^{-3}) / (8.31 	imes 290) \approx 0.485 \; kg$.Mass of oxygen taken out $\Delta m = m_{1} - m_{2} = 0.624 - 0.485 = 0.139 \; kg \approx 0.13 \; kg$.

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