[AdiDex]

Doubt No. 1
I was watching the video lecture from MIT 5.60 , In 4rth lecture I found this

"so we're going to define enthalpy as u + pV, these are all functions of state here, So H is a function of state,
and we're going to see now how this is, indeed, related to the heat flow in and out of the system. If you have a
constant pressure, "RERVERSIBLE WORK PROCESS". Let's take a system. Under constant pressure T1, V1, going to a
second -- this is the system, so let me write the system here. And it's more dramatic if the system is a gas, p, T2,
V2, And let's look at what happens to these functions of state, to H to u under this transformation "

How can an isobaric process be a reversible process ?? It will not be at equilibrium with surrounding at any point before its final state ..!??
I don't know what does it mean ?? any help ??
 
Doubt No. 2
In 3rd lecture, I found this

"Adiabatic means there's no heat transferred in or out of the system. We don't
say anything about whether the system does work or has worked on -- implicitly here, we're talking about a closed
system, so there's no mass leaving the system. But if it's adiabatic, then dq is equal zero, and for an adiabatic process, then du is equal to dw.

OK, now this is a point we want to be careful. Be careful. Suppose we are doing an adiabatic process. We can do
it reversibly, or we can do it irreversibly. So suppose we do it irreversibly. Suppose we just remove stops, and the
system slams up against the other stops. Did no work. Does that mean du is zero? You bet it does not. So, it's
important to write this little thing on here, du is equal to the reversible work, not just the work. "

As far as I know Internal energy of a given Ideal gas only depends upon Temperature . Lets expand it against vacuum , so its state change from P₁ , V₁  P₂ , V₂ . So it's  Internal energy should not change as No heat is transferred and No work is done .
If we had done this reversibly then there should be some work done .

So i am confused now , Where I am going wrong .