Answer
(a) $81.73\times10^{-5}m^{3}$
(b) 2.28 g
Work Step by Step
(a) Let's apply the ideal gas law $PV=nRT$ to find the volume of the tank.
$PV=nRT=>V=\frac{nRT}{P}$ ; Let's plug known values into this equation.
$V=\frac{(\frac{11\space g}{70.9\space g/mol})(8.31\space J/mol\space K)(273+82)K}{5.6\times10^{5}Pa}=81.73\times10^{-5}m^{3}$
Volume of the tank = $81.73\times10^{-5}m^{3}$
(b) We can write,
Mass of chlorine gas = 11 g - mass of the gas remaining in the tank
$m_{cl}=11\space g-n_{R}(Molecular \space mass)$
From the ideal gas law, we can get $n=\frac{PV}{RT}$
$m_{cl}=11\space g-(\frac{PV}{RT})(70.9\space g/mol)$ ; Let's plug known values into this equation.
$m_{cl}=11\space g-\frac{(3.8\times10^{5}Pa)(8.17\times10^{-4}m^{3})}{(8.31\space J/mol\space K)(31+273)K}(70.9\space g/mol)=2.28\space g$
Mass of the chlorine gas = 2.28 g
