Here, initially in the oxygen cylinder Volume $V_1 = 30\,\mathrm{litres} = 30 \times 10^{-3}\,\mathrm{m}^3$Pressure $P_1 = 15\,\mathrm{atm}$$= 15 \times 1.013 \times 10^{5}\,\mathrm{Pa}$Temperature $T_1 = 27 + 273 = 300\,\mathrm{K}$If the cylinder contains $n_1$ moles of the oxygen gas then $P_1V_1 = n_1RT,$$n_{1}= \frac{P_{1} V_{1}}{R T_{1}}$$= \frac{15 \times 1.013 \times 10^{5} \times 30 \times 10^{-3}}{8.31 \times 300}$$=18.253$Initial mass of oxygen in the cylinder$m_{1}=n_{1} \times$ molecular weight of $\mathrm{O}_{2}$$=18.253 \times 32 = 584.1\,\mathrm{g}$Finally in the oxygen cylinder, let $n_2$ moles of oxygen be left Final volume $V_2 = 30 \times 10^{-3}\,\mathrm{m}^3$Final pressure $P_2 = 11\,\mathrm{atm}$$= 11 \times 1.013 \times 10^{5}\,\mathrm{Pa}$Final temperature $T_2 = 17 + 273 = 290\,\mathrm{K}$$n_2 = ?$$\text{Now}, P_2 V_2 = n_2 R T_2$$n_2 = \frac{P_2 V_2}{R T_2}$Therefore, final mass of the oxygen gas in the cylinder$m_2 = n_2 \times$ molecular weight of $\mathrm{O}_2 = 13.847 \times 32 = 443.1\,\mathrm{g}$$\therefore$ Mass of oxygen gas taken out $= m_1 - m_2$$= (584.1 - 443.1) = 141\,\mathrm{g}$