To solve this problem, we need to compare the work done during isothermal and adiabatic compression of a gas.$\\$For an isothermal process, the work done $W_{iso}$ is given by:$\\ W_{iso} = nRT \ln\left(\frac{V_i}{V_f}\right) \\$For an adiabatic process, the work done $W_{adi}$ is given by:$\\ W_{adi} = \frac{P_i V_i - P_f V_f}{\gamma - 1} \\$where $\gamma$ is the adiabatic index (ratio of specific heats), $\gamma = \frac{C_p}{C_v}. \\$Given that the volume is reduced to half in both processes:$\\ V_f = \frac{V_i}{2} \\$For the isothermal process:$\\ W_{iso} = nRT \ln(2) \\$For the adiabatic process, we use the relation:$\\ P_i V_i^{\gamma} = P_f V_f^{\gamma} \\$Substituting $V_f = \frac{V_i}{2},$ we get:$\\ P_f = P_i \left(\frac{V_i}{V_f}\right)^{\gamma} = P_i (2)^{\gamma} \\$Substitute $P_f$ into the work done for adiabatic process:$\\ W_{adi} = \frac{P_i V_i - P_i (2)^{\gamma} \frac{V_i}{2}}{\gamma - 1} \\ W_{adi} = \frac{P_i V_i \left(1 - \frac{(2)^{\gamma}}{2}\right)}{\gamma - 1} \\$Since $P_i V_i = nRT$ for an ideal gas, we have:$\\ W_{adi} = \frac{nRT \left(1 - \frac{(2)^{\gamma}}{2}\right)}{\gamma - 1} \\$Comparing $W_{iso}$ and $W_{adi},$ we observe:$\\$• $W_{iso} = nRT \ln(2) \\$• $W_{adi} = \frac{nRT \left(1 - \frac{(2)^{\gamma}}{2}\right)}{\gamma - 1} \\$Since $\ln(2) \approx 0.693$ and $\frac{(2)^{\gamma}}{2}$ is greater than 1 for $\gamma > 1, \\$the term $\left(1 - \frac{(2)^{\gamma}}{2}\right)$ is negative, making $W_{adi}$ larger in magnitude.$\\$Thus, compressing the gas through an adiabatic process requires more work.$\\$Therefore, the correct option is 4: Compressing the gas through adiabatic process will require more work to be done.