As depicted in the figure, an ideal gas transitions from state i to...

From the given $p-T$ graph, the process is represented by a straight line passing through the origin.

This implies that the pressure $p$ is directly proportional to the temperature $T$, i.e., $p \propto T$, or $\dfrac{p}{T} = \text{constant}$.

According to the ideal gas equation, $pV = nRT$, which can be rewritten as $\dfrac{p}{T} = \dfrac{nR}{V}$.

Since $\dfrac{p}{T}$ is constant and $nR$ is constant for a given amount of gas, the volume $V$ must be constant. Thus, the process is isochoric.

The work done by a gas is given by $W = \int p \, dV$.

Since the volume is constant, the change in volume $dV = 0$.

Therefore, the work done by the gas during this process is zero.