Answer
See the graph below.
Work Step by Step
The given figure shows that the gas undergoes a compression-adiabatic process.
In this process,
$$\boxed{Q=0}$$
since no heat comes in or comes out of the system, it is perfectly isolated.
The work done is positive since $dV$ is negative because the volume of the gas decreases.
$$\boxed{W\gt 0}$$
We can see, from the given graph, that
$$\Delta T\gt 0$$
which means that the final temperature of the system is greater than its initial temperature.
Thus,
$$\boxed{\Delta T\gt 0}$$
See the bar-charts graph of this process below.
