Relation between enthalpy and heat for a non-ideal gas 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?
 A: Your assessment for an open system operating at steady state is correct provided that you add or remove the exact amount of heat necessary for the exit pressure of the gas to exactly match its inlet pressure.  The gas not being an ideal gas is irrelevent.
The change in enthalpy being equal to the heat added is correct only for an isobaric process in a closed system.
A: $C$ is defined as follows:
$$C = \frac 1n \frac {dQ}{dT}$$
Therefore heat supplied at constant pressure is given as:
$$Q=nC_p\Delta T$$
Where $C_p$ is just a special symbol used here to specify the given case of constant pressure.
So your problem is solved here.
But something does change. And guess what, it's Enthalpy.
For an ideal gas enthalpy is only dependent on temperature and hence its value is equal to heat supplied at the constant pressure.
But for a non-ideal gases it does depend on the pressure of the gas and hence Enthalpy is change not equal to the heat supplied at constant pressure. Rather it depends as follows:
$$ \Delta H = Q_p + \int \left (\frac {\partial H}{\partial P} \right )_T dP$$
