A mixture of 1 mol of a monatomic and 1 mol of a rigid diatomic ideal gas is slowly taken through a process in which the pressure is directly…

For the mixture (2 mol in all) the total heat capacity at constant volume is $C_V=\left(\tfrac32+\tfrac52\right)R=4R$, so the internal energy change is $\Delta U=4R\,\Delta T$. In the process $P=kV$ the work is $W=\int P\,dV=\tfrac{k}{2}(V_2^2-V_1^2)=\tfrac12(P_2V_2-P_1V_1)$. Using $PV=nRT$ with $n=2$ this becomes $W=\tfrac12\,(2R)\Delta T=R\,\Delta T$. The first law gives $Q=\Delta U+W=4R\Delta T+R\Delta T=5R\Delta T$. Therefore $Q/W=5$. Note that the answer does not depend on how large the temperature change is, only on the gas composition, because both $\Delta U$ and $W$ are proportional to $\Delta T$. A mistake is to use the molar $C_P$ or to count only one mole.