One mole of a monatomic ideal gas undergoes a quasi-static process described by P=P0(4-V/V0) as its volume increases from V0 to 3V0. The gas absorbs…

Let $x=V/V_0$. For a monatomic gas $dU=\tfrac32\,d(PV)$ and $dW=P\,dV$, so $dQ=\tfrac32\,d(PV)+P\,dV$. Here $PV=P_0V_0\,(4x-x^2)$, so $d(PV)=P_0V_0(4-2x)\,dx$ and $P\,dV=P_0V_0(4-x)\,dx$. Therefore $dQ=P_0V_0\left[\tfrac32(4-2x)+(4-x)\right]dx=P_0V_0\,(10-4x)\,dx$. Heat is absorbed while $10-4x>0$, i.e. up to $x=\tfrac52$, beyond which $dQ<0$ and heat is rejected. The net heat absorbed from $x=1$ to $x=\tfrac52$ is $\tfrac92P_0V_0$. Note that $x=2$ (option A) is where $PV$, and hence the temperature, is maximum; the gas still absorbs heat after that point because the work done exceeds the fall in internal energy. At $x=3$ the process ends, but heat is already being rejected before it. The correct option is (B).