Answer
See the detailed answer below.
Work Step by Step
The given figure shows that the volume is constant while the pressure is decreasing, so it is an isochoric process.
In this process, the work done is zero since the volume is constant.
$$\boxed{W=0}$$
We know, for an ideal gas, that
$$\dfrac{P_1 \color{red}{\bf\not} V_1}{T_1}=\dfrac{P_2 \color{red}{\bf\not} V_2}{T_2}$$
Hence,
$$\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}$$
and we can see, in the given graph, that $P_1\gt P_2$, so
$$\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\gt 1$$
Hence,
$$ T_1\gt T_2 $$
$$\boxed{\Delta T\lt 0}$$
This means that the temperature decreases during this process which means that the system loses some heat.
Thus,
$$\boxed{Q\lt 0}$$
See the bar-charts graph of this process below.
