One mole of a diatomic ideal gas (rotational but no vibrational modes active) undergoes a process in which its pressure is directly proportional to…

With $P=\alpha V$ and $PV=RT$ we get $T\propto V^2$, so doubling the volume quadruples the temperature: $\Delta T=3T_0$, where $T_0=P_0V_0/R$. The work done is $W=\int_{V_0}^{2V_0}\alpha V\,dV=\tfrac32\alpha V_0^2=\tfrac32 P_0V_0=\tfrac32RT_0$. The internal energy change for a diatomic gas is $\Delta U=\tfrac52R\Delta T=\tfrac{15}{2}RT_0$. By the first law, $Q=\Delta U+W=\tfrac{15}{2}RT_0+\tfrac32RT_0=9RT_0$. Therefore $W/Q=\tfrac{3/2}{9}=\tfrac16$. Equivalently the molar heat capacity along this path is $C_V+R/2=3R$, giving $Q=3R\cdot3T_0=9RT_0$. As a check, write $dQ=C_VdT+P\,dV$. Since $T\propto V^2$, $dT/T=2\,dV/V$, so $P\,dV=RT\,dV/V=\tfrac12R\,dT$, and the molar heat capacity of the path is $C=\tfrac52R+\tfrac12R=3R$, which again gives $Q=3R\cdot3T_0=9RT_0$. Thus exactly one sixth of the heat supplied appears as work, and the other five sixths raise the internal energy of the gas.