The work done by a gas in a thermodynamic process corresponds to the area under the process curve on a P-V diagram, given by $W = \int P \, dV$. If the process is isochoric (constant volume), the change in volume $dV = 0$, resulting in zero work done. Since the P-V diagram is missing from the provided text, we cannot visually confirm the exact nature of path $bc$. However, assuming path $bc$ represents an isochoric process as suggested by the probable answer, the work done along this path would be zero.