The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA.  The part ABC is a semi circle and CDA is half of an ellipse. Then

P-V graph is not rectangular hyperbola. Therefore, process A - B is not isothermal.

In process BCD, product of pV (therefore temperature and internal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence, Q will be negative or heat will flow out of the gas.

${\mathrm{W}}_{ABC}=\text{ positive }$

For clockwise cycle on p-V diagram with P on y-axis, net work done is positive.