Enthelpy Transformation: P=Constant, ΔH=Q?

[Elio Gruttadauria]

In a trasformation in which P=costant, but internal pressure is different from external pressure, ΔH=Q?

I'm asking this question because I know that
Q=ΔU+PΔV (where P is the external pressure)

and
H=U+PV (where P is the sistem pression, so the internal pressure)

Am I right?

 

[Chestermiller]

You are talking about an irreverisble process in which you suddenly increase or decrease the external pressure, and then allow the system re-equilibrate, correct? In such a process, three things are happening that you may not be aware of:
1. The pressure and temperature inside the system are not spatially uniform, and vary from location to location. So, it is not possible to identify a specific value of the pressure that characterizes the "system pressure."
2. The contents of the system is deforming rapidly, and this contributes viscous stresses to the force on the piston, such that the force per unit area doesn't just depend on the volume change but also on the rate of change. This rate effect is not present in a reversible process.
3. For a frictionless, massless piston, the force per unit area at the internal piston face (which includes viscous stresses) exactly matches the external force per unit area at the external piston face. This is why the work done by the system is equal to the work done on the surroundings.

Even though we say that an irreversible process is at constant pressure, what we really mean is that the external pressure is constant after we suddenly change the external pressure to a new value at time zero. But this pressure is not equal to the value of the system pressure prior to time zero. And the pressure of the system between the initial and final equilibrium states does change. In the equation for enthalpy, we are referring to the values of the parameters in an equilibrium state.