One mole of a monatomic ideal gas is taken quasi-statically through a process in which its pressure is proportional to its volume, P=α V, until the…

For $P=\alpha V$ we have $PV=\alpha V^2=RT$, so $T\propto V^2$ and the temperature rises by a factor 4. With $V_0\to 2V_0$ and $P_0=\alpha V_0$, the temperature change is $\Delta T=3P_0V_0/R$. The work is $W=\int \alpha V\,dV=\tfrac32\alpha V_0^2=\tfrac32 P_0V_0$. The internal energy change is $\Delta U=\tfrac32 R\Delta T=\tfrac92P_0V_0$, so $Q=\Delta U+W=6P_0V_0$. Equivalently the molar heat capacity of this process is $C=C_v+R/2=2R$, giving the same $Q=2R\Delta T=6P_0V_0$. Hence $W/Q=1.5/6=1/4$. Note that the process is neither isobaric nor isothermal, so the heat capacity must be derived from the process equation rather than recalled. Using $C_p=\tfrac52R$ by mistake gives $1/5$, while taking $Q=\Delta U$ gives $1/3$.