Answer
See the figure below.
Work Step by Step
The given figure shows that the pressure is constant while the volume is expanding, so it is an expansion-isobaric process.
In this process,
$$\dfrac{ \color{red}{\bf\not} P_1V_1}{T_1}=\dfrac{ \color{red}{\bf\not} P_2V_2}{T_2} $$
Thus,
$$\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}$$
and since the $V_2\gt V_1$, so
$$\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}\gt 0$$
Hence,
$$ \dfrac{T_2}{T_1}\gt 0$$
$$\boxed{T_2\gt T_1}$$
This means that the system gains some energy from the surroundings since its final temperature is greater than the initial one.
$$\boxed{Q\gt 0}$$
Also, $V_2\gt V_1$, so the work is negative.
$$\boxed{W\lt 0}$$
