Is the work done by gas necessarily positive if the cycle is travelled clockwise in the $P-V$ plane (and viceversa)? Is it always true, for any cycle followed by a gas, that, if the cycle is "travelled" clockwise in the $P-V$ plane then the work exchanged by the gas is positive, and viceversa for the clockwise direction?
I came up with this situation, which I represented in the picture. The irreversible cycle is made of an isothermal compression, and two adiabatics. One of the adiabatics must be irreversible for the cycle to exists. 

The cycle seems possible to me because since $$\Delta S_{universe}=\Delta S_{environment}=\Delta S_{environment, A->B}>0$$
This cycle is clockwise, but $$Q_{cycle}=W_{cycle}=Q_{A->B}<0$$
So this appears in contrast with the rule
$$\mathrm{clockwise} \implies W>0$$
$$\mathrm{anticlockwise} \implies W<0$$
Is that possible or am I missing something?
Sign convention:

 A: You have to be careful when drawing irreversible transformations on the PV plane. Apart from the fact that drawing them is meaningless because they are not a set of equilibrium points, you cannot just assign some arbitrary property to them.
You drew the blue curve and called it an "irreversible adiabatic", giving it some arbitrary slope. Fine, since we don't know what the "slope" of an irreversible adiabatic should be. But what if I drew another "irreversible adiabatic" connecting A and B? 
We would then have 
$$\Delta U = 0 \to Q = W$$ 
But $Q=0$ since every process is adiabatic, so that we would obtain
$$W=0$$
which is obviously wrong. 
A: The area enclosed within the loop is the net work and for the complete cycle you've illustrated, work (energy) flows both into and out of the system at different points within the cycle. But there is a net loss.
From $A$ to $B$ and $B$ to $C$, the direction is positive (by right hand convention) and so energy flows into the system. From $C$ to $A$ the direction is negative, and so energy flows out of the system.
Since integration along $V$ is negative along the direct path from $C$ to $A$ and greater in magnitude than the integration under the path from $A$ to $B$ then $B$ to $C$
(which is positive), then the net energy flow and work is negative.
But this can be directly determined by just integrating the area within the path and observing the path moves in a clockwise direction.
