Is enthalpy defined under non-isobaric conditions? My question is this,
Does Enthalpy have a meaning under non-isobaric conditions?
Is its existence as a property of a system independent of whether the system is under isobaric condition or not?
Edit:- I wanted to know if enthalpy as a property of a system is valid only if from its creation to current state, it undergoes only isobaric processes.
 A: Enthalpy is just a measure of the total energy of a thermodynamic system, so it always exists.
I'm not sure what you mean by non-isobaric conditions. Do you mean a system that isn't the same pressure everywhere (and therefore isn't in equilibrium) or do you mean a change during which the pressure changes. In both cases the enthalpy of the whole system has a definite value, though it may not be easy to calculate.
A: The question is rather vaguely formulated but if you mean "is there a systematic method for computing the enthalpy from the equations of state?", then the answer is yes.  For if these are $$T=f(p,V), \quad S=g(p,V)  $$
for suitable functions $f$ and $g$ of $p$ and $V$, then it follows from the defining condition $dH=TdS + V dp$ for the enthalpy $H$ and an elementary application of the inverse function theorem and the chain rule that if we regard $H$ as a function of $p$ and $V$, then
$$dH = (V+f g_1)dp + fg_2 dV.$$ (We are using subscript to denote partial derivatives with respect to $p$ and $V$).  This equation can then be solved by the standard methods for  exact differential equations (that the equation is exact is a consequence of the Maxwell relations). For relatively simple models, this can be done by hand---otherwise there are standard numerical methods available.  For example,  a two-line calculation shows that $H=\dfrac \gamma{\gamma-1} T$ for the ideal gas, which we assume to have equations of state $T=pV$ and $S=\dfrac 1{\gamma-1} \ln p + \dfrac \gamma{\gamma-1} \ln V$. In the case of the van der Waals gas (in the simplified version $$T=\left( p + \frac 1 {V^2}\right ) \left (V - 1\right ), \quad S = \frac 1 {\gamma-1}\ln \left (p + \frac 1 {V^2}\right )+ \frac \gamma{\gamma-1} \ln (V-1)),$$
the calculation is a bit messier but can be carried out by hand in a few lines (this is less work than writing down the answer in TeX).
A: Enthalpy is as follows:
$$dH=TdS+VdP$$
You can get that from the famous diagrams "Good physicists have studied under very fine teachers", (it's the best mnemotechnique rule to remember all the relations of potentials/functions).
Under isobaric conditions $dP=0$ and $dH=TdS$, but if the pressure is not constant during a proccess, then enthalpy is also defined, just as before: $dH=TdS+VdP$.
