The temperature is\(T = 295\;{\rm{K}}\(.

The gauge pressure is \(P = 13800\;{\rm{kPa}}\).

The volume is \(V = 0.014\;{{\rm{m}}^3}\).

The atmospheric pressure is \({P_{\rm{a}}} = 1\;{\rm{atm}}\).

The flow rate is \(c = 2.1\;{\rm{L/min}}\).

In this problem, first, calculate the volume occupied at atmospheric pressure as the oxygen pressure drops. Consider that the final pressure inside the cylinder is the atmospheric pressure.

The relation from the ideal gas law can be written as:

\(\begin{aligned}{c}PV &= P'V'\\V' &= \left( {\frac{{PV}}{{{P_{\rm{a}}}}}} \right)\end{aligned}\(

Here,\(V'\(is the volume at atmospheric pressure.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}V' &= \left( {\frac{{\left( {13800 \times {{10}^3}\;{\rm{Pa}} + 1.013 \times {{10}^5}\;{\rm{Pa}}} \right)}}{{\left( {1\;{\rm{atm}} \times \frac{{1.013 \times {{10}^5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}}} \right)}}} \right(\left( {0.014\;{{\rm{m}}^3}} \right)\\V' &= \left( {1.921\;{{\rm{m}}^3} \times \frac{{1000\;{\rm{L}}}}{{1\;{{\rm{m}}^3}}}} \right)\\V' &= 1921\;{\rm{L}}\end{aligned}\(

Since 14 L of gas is not available because it is left in the container, the total available volume is:

\(\begin{aligned}{l}{V_0} &= V' - V\\{V_0} &= 1921\;{\rm{L}} - 14\;{\rm{L}}\\{V_0} &= 1907\;{\rm{L}}\end{aligned}\)

The relation to find the time can be written as:

\(\begin{aligned}{l}t &= \frac{{{V_0}}}{c}\\t &= \left( {1907\;{\rm{L}}} \right)\left( {\frac{{1\;{\rm{min}}}}{{2.1\;{\rm{L}}}}} \right)\\t &= 908.08\;{\rm{min}}\end{aligned}\)

Thus, \(t = 908.08\;{\rm{min}}\) is the required time.