Thermodynamics – How to Interpret the Definition and Differential Relation of Enthalpy

Question

I know that enthalpy is defined as $H=E+PV$. So for a system the enthalpy is the sum of internal energy and the amount of work needed at constant pressure $P$ to give it's volume $V$. But the differential of enthalpy is given by legendre transformation,
$$dH=TdS+VdP$$
For an adiabatic system ($dQ=0$) if work is done on the system at a constant pressure the mechanical equilibrium is achived when enthalpy is minimized. So whenever there is some work done on the system at constant pressure with no heat exchange, $dH$ is negative.

So my question is, if $dQ=0$, then $TdS$ is also zero and at constant pressure $dP=0$. So by the definion of enthalpy, $dH$ is not zero it's negative, but from the differential $dH=0$ always. Can anyone explain how do I interpret this?

Answer 1

For a reversible adiabatic (constant entropy) process for an ideal gas, pressure varies with volume according to

$$PV^{\gamma}=C$$

Where $C$ is a constant. So pressure cannot be constant.

A constant pressure adiabatic process is an irreversible process. See:

https://physics.stackexchange.com/questions/577884/is-a-constant-pressure-adiabatic-irreversible-expansion-possible#:~:text=That%2C%20of%20course%2C%20can',external%20pressure%20expansion%20or%20contraction.

So a constant pressure adiabatic process generates entropy, i.e., $dS>0$, in spite of the fact that $Q=0$.

Hope this helps.