[Dev]

 Hi. Is there a physical significance to the quantity of enthalpy similar to internal energy?
Is there any physical significance of PV (Pressure X Volume)
 

And why is work done by a system equal to (Pext)dV (Pext = external pressure)
I mean how does the gas inside the system manage to not apply P = nRT/V but instead it applies Pext)

Also, is dSt (total change in entropy that is system + surroundings) a state function?

[Soumalya]

Hello,
       To understand what enthalpy actually is let us start by analyzing it's mathematical interpretation which is H=U+PV.

"U" refers to the internal energy of a system whereas "P" refers to the externally applied pressure(which is in balance with the internal pressure of the system at all the thermodynamic equilibrium states of a process) and "V" the volume occupied by the system.
 Now, for a system to undergo a process there must be change in its thermodynamic properties such that change in enthalpy could be defined as, ΔH=ΔU+Δ(PV).

Upon expansion, Δ(PV)=PΔV+VΔP

Again PΔV is the work of expansion for a constant pressure non flow process whereas VΔP is the work for a flow process within a constant control volume.

Thus enthalpy could be defined as the summation of 'internal energy' of a system plus energy in terms of the work done in creating the system against some external pressure or the surroundings.The 'PV' term implies the work done to create the system in existence against the surroundings.One may ask why is it only 'PV' not Δ(PV) between two states?

Because the value of enthalpy is assigned a zero value at certain standard conditions so the change in enthalpy is ΔH=(U-0)+(PV-0) when the system undergoes a change from the standard state of zero enthalpy to a state of pressure "P" and volume "V".