To solve this problem, we need to find the ratio of work done by the gas to the heat absorbed by it.$\\$Given: The process is from state A to state B, and the graph shows a linear relationship between $V$ and $T. \\$For a monatomic ideal gas, the work done $(W)$ during an expansion is given by:$\\ W = \int PdV \\$Using the ideal gas law $PV = nRT,$ we can express $P$ as:$\\ P = \frac{nRT}{V} \\$Substitute $P$ in the work done equation:$\\ W = \int \frac{nRT}{V} dV \\$Since $V$ is directly proportional to $T,$ we can write $V = kT,$ where $k$ is a constant.$\\$Thus, $W = \int \frac{nRT}{kT} d(kT) = nR \int dT = nR(T_B - T_A) \\$The heat absorbed $(Q)$ by the gas is given by the first law of thermodynamics:$\\ Q = \Delta U + W \\$For a monatomic ideal gas, the change in internal energy $(\Delta U)$ is:$\\ \Delta U = \frac{3}{2}nR(T_B - T_A) \\$Substitute $\Delta U$ and $W$ into the equation for $Q: \\ Q = \frac{3}{2}nR(T_B - T_A) + nR(T_B - T_A) \\ Q = \frac{5}{2}nR(T_B - T_A) \\$Now, find the ratio of work done to heat absorbed:$\\ \frac{W}{Q} = \frac{nR(T_B - T_A)}{\frac{5}{2}nR(T_B - T_A)} \\ \frac{W}{Q} = \frac{2}{5} \\$Therefore, the ratio of work done by the gas to the heat absorbed by it is $\frac{2}{5}. \\$This corresponds to Option 3.