# Enthalpy at constant pressure/volume

Enthalpy is defined as the heat the goes into (leaves) a system at constant pressure.
$$dH =TdS+ d(PV) = TdS + PdV + VdP$$

For some reason, most texts simply write $$dH = TdS + VdP$$ but haven't they cancelled out the wrong term? Under isobaric conditions $$dP = 0$$ hence the equation should read $$dH = TdS + PdV$$. Why isn't this the case?

Enthalpy is defined as $$H=U+pV$$ irrespective of what process the system is subjected to. It, enthalpy $$H$$, is a state function and in equilibrium thermostatics its definition has nothing to do with the process. There is process and there is state. When $$p$$ is constant, $$dp=0$$, then of course $$dH=dU+pdV$$. If the system can be described by two variables, such as $$dU=TdS-pdV$$ then and only then $$dH|_{dp=0}=TdS.$$ Otherwise you may have some other work variables, such electrostatic or magnetostatic, etc., whose work is represented by $$\delta w$$ and you have in general $$dU=TdS-pdV+\delta w$$ with $$dH_{dp=0}=TdS+\delta w.$$
• yes and no. The exact differential is an exact differential if it is taken between two equilibrium states, it does not matter what the process is that connects them. If the process is reversible then external parameters determine the internal parameters $dH$ or $dU$ or $dS$, etc., and can be measured by observing the externally supplied quantities and work. Commented Jul 20 at 23:10