To solve this problem, we need to analyze the two processes: isothermal expansion and adiabatic expansion.$\\$1. Isothermal Expansion:$\\$• The gas expands isothermally from volume $V$ to $2V. \\$• For an isothermal process, the pressure-volume relationship is given by $PV =$constant.$\\$Let $P_1$ be the initial pressure and $P_2$ be the pressure after isothermal expansion.$\\ P_1 V = P_2 (2V) \\ P_2 = \frac{P_1 V}{2V} = \frac{P_1}{2} \\$2. Adiabatic Expansion:$\\$• The gas then expands adiabatically from volume $2V$ to $16V. \\$• For an adiabatic process, the relationship is $P V^\gamma =$constant, where $\gamma = \frac{5}{3}. \\$Let $P_3$ be the final pressure after adiabatic expansion.$\\ P_2 (2V)^\gamma = P_3 (16V)^\gamma \\$Substitute $P_2 = \frac{P_1}{2}: \\ \left(\frac{P_1}{2}\right) (2V)^\gamma = P_3 (16V)^\gamma \\$Simplify the equation:$\\ \frac{P_1}{2} \cdot (2V)^{\frac{5}{3}} = P_3 \cdot (16V)^{\frac{5}{3}} \\$Since $(16V) = (2^4 V),$ we have:$\\ \frac{P_1}{2} \cdot (2^{\frac{5}{3}} V^{\frac{5}{3}}) = P_3 \cdot (2^{4 \cdot \frac{5}{3}} V^{\frac{5}{3}}) \\$Cancel $V^{\frac{5}{3}}$ from both sides:$\\ \frac{P_1}{2} \cdot 2^{\frac{5}{3}} = P_3 \cdot 2^{\frac{20}{3}} \\$Rearrange to solve for $P_3: \\ P_3 = \frac{P_1}{2} \cdot \frac{2^{\frac{5}{3}}}{2^{\frac{20}{3}}} \\ P_3 = \frac{P_1}{2} \cdot 2^{\frac{5}{3} - \frac{20}{3}} \\ P_3 = \frac{P_1}{2} \cdot 2^{-\frac{15}{3}} \\ P_3 = \frac{P_1}{2} \cdot 2^{-5} \\ P_3 = \frac{P_1}{2 \cdot 32} \\ P_3 = \frac{P_1}{64} \\$Therefore, the final pressure of the gas is $\frac{P}{64}. \\$This corresponds to Option 3.