Consider a non-ideal gas passing through an isobaric process, with $C_p$ not affected by temperature. Would $C_p\Delta = Q$? Where $Q$ is the heat the gas received?
I think it is not.
Here is my reasoning, $$dH = C_p dT + V(1-\alpha T)dP \\ \implies dH = C_pdT $$ But $$\Delta H = W_s + Q$$ where $W_s$ is shaft work. So $$Q = C_p \Delta T - W_s,$$ which is not the statement we have been given.
Is this an accurate logical proof? Or is shaft work zero for an isobaric process?