# 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.

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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).

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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.

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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: $dH=TdS+VdP$.

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