thermodynamics - Ideal gas reversible cycle

So I got this cycle on a quiz. As of my understanding, in the transformation 1-2 there is also received heat and given heat ( sorry if that's not the english exprimation, not my first language). Now I have a question:

As I understand it, there is a given point M on 1-2 until which heat is received, and from M to 2 it is given. You can solve for it with the general equation p=aV + b. Anyways, why is it imperative for the received heat to be first and not the other way around? Is it because of the 2nd law of thermodynamics? And if so, could you explain it for this case?

Thank you in advance!

Answer

No. It's the result of applying the first law of thermodynamics.

The equation for the straight line between the initial and final volumes is

$$P=P_0\left(3-\frac{V}{V_0}\right)\tag{1}$$

From the 1st law of thermodynamics for this reversible process, we have:

$$dU=nC_vdT=dQ-PdV$$ or $$dQ=nC_vdT+PdV\tag{2}$$ Combining this with the ideal gas law gives: $$dQ=\frac{C_v}{R}d(PV)+PdV\tag{3}$$ Substituting Eqn. 1 into Eqn. 3 gives: $$dQ=\frac{C_v}{R}P_0\left(3-2\frac{V}{V_0}\right)dV+P_0\left(3-\frac{V}{V_0}\right)dV\tag{4}$$ At $V=V_0$, this becomes $$dQ=P_0\left(\frac{C_v}{R}+2\right)dV$$ which is greater than zero. At $V=2V_0$, Eqn. 4 yields: $$dQ=P_0\left(1-\frac{C_v}{R}\right)dV$$ Since $C_v/R>1$, dQ is negative at $V=2V_0$. So dQ changes sign between the initial volume and the final volume.